Algebra Examples

$y = x^{2} - 2 x - 5$ , $y = x^{3} - 2 x^{2} - 5 x - 9$

Eliminate the equal sides of each equation and combine.

$x^{2} - 2 x - 5 = x^{3} - 2 x^{2} - 5 x - 9$

Solve $x^{2} - 2 x - 5 = x^{3} - 2 x^{2} - 5 x - 9$ for $x$ .

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Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{3} - 2 x^{2} - 5 x - 9 = x^{2} - 2 x - 5$

Move all terms containing $x$ to the left side of the equation.

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Subtract $x^{2}$ from both sides of the equation.

$x^{3} - 2 x^{2} - 5 x - 9 - x^{2} = - 2 x - 5$

Add $2 x$ to both sides of the equation.

$x^{3} - 2 x^{2} - 5 x - 9 - x^{2} + 2 x = - 5$

Subtract $x^{2}$ from $- 2 x^{2}$ .

$x^{3} - 3 x^{2} - 5 x - 9 + 2 x = - 5$

Add $- 5 x$ and $2 x$ .

$x^{3} - 3 x^{2} - 3 x - 9 = - 5$

Add $5$ to both sides of the equation.

$x^{3} - 3 x^{2} - 3 x - 9 + 5 = 0$

Add $- 9$ and $5$ .

$x^{3} - 3 x^{2} - 3 x - 4 = 0$

Factor $x^{3} - 3 x^{2} - 3 x - 4$ using the rational roots test.

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If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 4, \pm 2$

$q = \pm 1 + y = x^{3} - 2 x^{2} - 5 x - 9$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 4, \pm 2 + y = x^{3} - 2 x^{2} - 5 x - 9$

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

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Substitute $4$ into the polynomial.

$4^{3} - 3 \cdot 4^{2} - 3 \cdot 4 - 4 + y = x^{3} - 2 x^{2} - 5 x - 9$

Raise $4$ to the power of $3$ .

$64 - 3 \cdot 4^{2} - 3 \cdot 4 - 4 + y = x^{3} - 2 x^{2} - 5 x - 9$

Raise $4$ to the power of $2$ .

$64 - 3 \cdot 16 - 3 \cdot 4 - 4 + y = x^{3} - 2 x^{2} - 5 x - 9$

Multiply $- 3$ by $16$ .

$64 - 48 - 3 \cdot 4 - 4 + y = x^{3} - 2 x^{2} - 5 x - 9$

Subtract $48$ from $64$ .

$16 - 3 \cdot 4 - 4 + y = x^{3} - 2 x^{2} - 5 x - 9$

Multiply $- 3$ by $4$ .

$16 - 12 - 4 + y = x^{3} - 2 x^{2} - 5 x - 9$

Subtract $12$ from $16$ .

$4 - 4 + y = x^{3} - 2 x^{2} - 5 x - 9$

Subtract $4$ from $4$ .

$0 + y = x^{3} - 2 x^{2} - 5 x - 9$

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 3 x^{2} - 3 x - 4}{x - 4} + y = x^{3} - 2 x^{2} - 5 x - 9$

Divide $x^{3} - 3 x^{2} - 3 x - 4$ by $x - 4$ .

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Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$4$

$x^{3}$

$3 x^{2}$

$3 x$

$4$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			+	$x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 4 x^{2}$

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					+	$x^{2}$	-	$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} - 4 x$

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					-	$x^{2}$	+	$4 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					-	$x^{2}$	+	$4 x$
							+	$x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					-	$x^{2}$	+	$4 x$
							+	$x$	-	$4$

Divide the highest order term in the dividend $x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$x$	+	$1$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					-	$x^{2}$	+	$4 x$
							+	$x$	-	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	+	$1$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					-	$x^{2}$	+	$4 x$
							+	$x$	-	$4$
							+	$x$	-	$4$

The expression needs to be subtracted from the dividend, so change all the signs in $x - 4$

				$x^{2}$	+	$x$	+	$1$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					-	$x^{2}$	+	$4 x$
							+	$x$	-	$4$
							-	$x$	+	$4$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	+	$x$	+	$1$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$3 x$	-	$4$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$3 x$
					-	$x^{2}$	+	$4 x$
							+	$x$	-	$4$
							-	$x$	+	$4$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$x^{2} + x + 1 + y = x^{3} - 2 x^{2} - 5 x - 9$

Write $x^{3} - 3 x^{2} - 3 x - 4$ as a set of factors.

$(x - 4) (x^{2} + x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x^{2} + x + 1 = 0 + y = x^{3} - 2 x^{2} - 5 x - 9$

Set $x - 4$ equal to $0$ and solve for $x$ .

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Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x^{2} + x + 1$ equal to $0$ and solve for $x$ .

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Set $x^{2} + x + 1$ equal to $0$ .

$x^{2} + x + 1 = 0$

Solve $x^{2} + x + 1 = 0$ for $x$ .

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Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a} + y = x^{3} - 2 x^{2} - 5 x - 9$

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1} + y = x^{3} - 2 x^{2} - 5 x - 9$

Simplify.

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Simplify the numerator.

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One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{- 1 + i \sqrt{3}}{2}$

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 + i \sqrt{3}}{2}$

Factor $- 1$ out of $i \sqrt{3}$ .

$x = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$x = \frac{- 1 (1 - i \sqrt{3})}{2}$

Move the negative in front of the fraction.

$x = - \frac{1 - i \sqrt{3}}{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{- 1 - i \sqrt{3}}{2}$

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 - i \sqrt{3}}{2}$

Factor $- 1$ out of $- i \sqrt{3}$ .

$x = \frac{- 1 \cdot 1 - (i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$x = \frac{- 1 (1 + i \sqrt{3})}{2}$

Move the negative in front of the fraction.

$x = - \frac{1 + i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

The final solution is all the values that make $(x - 4) (x^{2} + x + 1) = 0$ true.

$x = 4, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Evaluate $y$ when $x = 4$ .

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Substitute $4$ for $x$ .

$y = {(4)}^{3} - 2 {(4)}^{2} - 5 \cdot 4 - 9$

Substitute $4$ for $x$ in $y = {(4)}^{3} - 2 {(4)}^{2} - 5 \cdot 4 - 9$ and solve for $y$ .

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Remove parentheses.

$y = 4^{3} - 2 {(4)}^{2} - 5 \cdot 4 - 9$

Remove parentheses.

$y = 4^{3} - 2 \cdot 4^{2} - 5 \cdot 4 - 9$

Remove parentheses.

$y = {(4)}^{3} - 2 {(4)}^{2} - 5 \cdot 4 - 9$

Simplify ${(4)}^{3} - 2 {(4)}^{2} - 5 \cdot 4 - 9$ .

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Simplify each term.

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Raise $4$ to the power of $3$ .

$y = 64 - 2 {(4)}^{2} - 5 \cdot 4 - 9$

Raise $4$ to the power of $2$ .

$y = 64 - 2 \cdot 16 - 5 \cdot 4 - 9$

Multiply $- 2$ by $16$ .

$y = 64 - 32 - 5 \cdot 4 - 9$

Multiply $- 5$ by $4$ .

$y = 64 - 32 - 20 - 9$

Simplify by subtracting numbers.

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Subtract $32$ from $64$ .

$y = 32 - 20 - 9$

Subtract $20$ from $32$ .

$y = 12 - 9$

Subtract $9$ from $12$ .

$y = 3$

Evaluate $y$ when $x = - \frac{1 - i \sqrt{3}}{2}$ .

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Substitute $- \frac{1 - i \sqrt{3}}{2}$ for $x$ .

$y = {(- \frac{1 - i \sqrt{3}}{2})}^{3} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Simplify ${(- \frac{1 - i \sqrt{3}}{2})}^{3} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$ .

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Simplify each term.

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Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1 - i \sqrt{3}}{2}$ .

$y = {(- 1)}^{3} {(\frac{1 - i \sqrt{3}}{2})}^{3} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the product rule to $\frac{1 - i \sqrt{3}}{2}$ .

$y = {(- 1)}^{3} \frac{{(1 - i \sqrt{3})}^{3}}{2^{3}} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $- 1$ to the power of $3$ .

$y = - \frac{{(1 - i \sqrt{3})}^{3}}{2^{3}} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $2$ to the power of $3$ .

$y = - \frac{{(1 - i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Use the Binomial Theorem.

$y = - \frac{1^{3} + 3 \cdot 1^{2} (- i \sqrt{3}) + 3 \cdot 1 {(- i \sqrt{3})}^{2} + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Simplify each term.

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One to any power is one.

$y = - \frac{1 + 3 \cdot 1^{2} (- i \sqrt{3}) + 3 \cdot 1 {(- i \sqrt{3})}^{2} + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

One to any power is one.

$y = - \frac{1 + 3 \cdot 1 (- i \sqrt{3}) + 3 \cdot 1 {(- i \sqrt{3})}^{2} + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $3$ by $1$ .

$y = - \frac{1 + 3 (- i \sqrt{3}) + 3 \cdot 1 {(- i \sqrt{3})}^{2} + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $3$ .

$y = - \frac{1 - 3 (i \sqrt{3}) + 3 \cdot 1 {(- i \sqrt{3})}^{2} + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $3$ by $1$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 {(- i \sqrt{3})}^{2} + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- i \sqrt{3}$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 ({(- i)}^{2} {\sqrt{3}}^{2}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the product rule to $- i$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 ({(- 1)}^{2} i^{2} {\sqrt{3}}^{2}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $- 1$ to the power of $2$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 (1 i^{2} {\sqrt{3}}^{2}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $i^{2}$ by $1$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 (i^{2} {\sqrt{3}}^{2}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite $i^{2}$ as $- 1$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 (- 1 {\sqrt{3}}^{2}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 (- 1 {(3^{\frac{1}{2}})}^{2}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 (- 1 \cdot 3^{\frac{1}{2} \cdot 2}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Combine $\frac{1}{2}$ and $2$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 (- 1 \cdot 3^{\frac{2}{2}}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Cancel the common factor of $2$ .

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Cancel the common factor.

$y = - \frac{1 - 3 i \sqrt{3} + 3 (- 1 \cdot 3^{\frac{2}{2}}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = - \frac{1 - 3 i \sqrt{3} + 3 (- 1 \cdot 3^{1}) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Evaluate the exponent.

$y = - \frac{1 - 3 i \sqrt{3} + 3 (- 1 \cdot 3) + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $3$ .

$y = - \frac{1 - 3 i \sqrt{3} + 3 \cdot - 3 + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $3$ by $- 3$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + {(- i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- i \sqrt{3}$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + {(- i)}^{3} {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the product rule to $- i$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + {(- 1)}^{3} i^{3} {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $- 1$ to the power of $3$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 - i^{3} {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Factor out $i^{2}$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 - (i^{2} \cdot i) {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite $i^{2}$ as $- 1$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 - (- 1 \cdot i) {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite $- 1 i$ as $- i$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 - - i {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $- 1$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + 1 i {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $i$ by $1$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + i {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite ${\sqrt{3}}^{3}$ as ${(3^{3})}^{\frac{1}{2}}$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + i \sqrt{3^{3}}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $3$ to the power of $3$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + i \sqrt{27}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite $27$ as $3^{2} \cdot 3$ .

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Factor $9$ out of $27$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + i \sqrt{9 (3)}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite $9$ as $3^{2}$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + i \sqrt{3^{2} \cdot 3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Pull terms out from under the radical.

$y = - \frac{1 - 3 i \sqrt{3} - 9 + i (3 \sqrt{3})}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Move $3$ to the left of $i$ .

$y = - \frac{1 - 3 i \sqrt{3} - 9 + 3 i \sqrt{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Subtract $9$ from $1$ .

$y = - \frac{- 3 i \sqrt{3} - 8 + 3 i \sqrt{3}}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Add $- 3 i \sqrt{3}$ and $3 i \sqrt{3}$ .

$y = - \frac{0 - 8}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Subtract $8$ from $0$ .

$y = - \frac{- 8}{8} - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Divide $- 8$ by $8$ .

$y = - - 1 - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $- 1$ .

$y = 1 - 2 {(- \frac{1 - i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1 - i \sqrt{3}}{2}$ .

$y = 1 - 2 ({(- 1)}^{2} {(\frac{1 - i \sqrt{3}}{2})}^{2}) - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the product rule to $\frac{1 - i \sqrt{3}}{2}$ .

$y = 1 - 2 ({(- 1)}^{2} \frac{{(1 - i \sqrt{3})}^{2}}{2^{2}}) - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $- 1$ to the power of $2$ .

$y = 1 - 2 (1 \frac{{(1 - i \sqrt{3})}^{2}}{2^{2}}) - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $\frac{{(1 - i \sqrt{3})}^{2}}{2^{2}}$ by $1$ .

$y = 1 - 2 \frac{{(1 - i \sqrt{3})}^{2}}{2^{2}} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $2$ to the power of $2$ .

$y = 1 - 2 \frac{{(1 - i \sqrt{3})}^{2}}{4} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Cancel the common factor of $2$ .

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Factor $2$ out of $- 2$ .

$y = 1 + 2 (- 1) \frac{{(1 - i \sqrt{3})}^{2}}{4} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Factor $2$ out of $4$ .

$y = 1 + 2 \cdot - 1 \frac{{(1 - i \sqrt{3})}^{2}}{2 \cdot 2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Cancel the common factor.

$y = 1 + 2 \cdot - 1 \frac{{(1 - i \sqrt{3})}^{2}}{2 \cdot 2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = 1 - 1 \frac{{(1 - i \sqrt{3})}^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite ${(1 - i \sqrt{3})}^{2}$ as $(1 - i \sqrt{3}) (1 - i \sqrt{3})$ .

$y = 1 - 1 \frac{(1 - i \sqrt{3}) (1 - i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Expand $(1 - i \sqrt{3}) (1 - i \sqrt{3})$ using the FOIL Method.

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Apply the distributive property.

$y = 1 - 1 \frac{1 (1 - i \sqrt{3}) - i \sqrt{3} (1 - i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the distributive property.

$y = 1 - 1 \frac{1 \cdot 1 + 1 (- i \sqrt{3}) - i \sqrt{3} (1 - i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the distributive property.

$y = 1 - 1 \frac{1 \cdot 1 + 1 (- i \sqrt{3}) - i \sqrt{3} \cdot 1 - i \sqrt{3} (- i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Simplify and combine like terms.

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Simplify each term.

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Multiply $1$ by $1$ .

$y = 1 - 1 \frac{1 + 1 (- i \sqrt{3}) - i \sqrt{3} \cdot 1 - i \sqrt{3} (- i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- i \sqrt{3}$ by $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} \cdot 1 - i \sqrt{3} (- i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} - i \sqrt{3} (- i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- i \sqrt{3} (- i \sqrt{3})$ .

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Multiply $- 1$ by $- 1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + 1 i \sqrt{3} (i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $\sqrt{3}$ by $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + \sqrt{3} i (i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $\sqrt{3}$ to the power of $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{1} \sqrt{3} i i}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $\sqrt{3}$ to the power of $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{1} {\sqrt{3}}^{1} i i}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{1 + 1} i i}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Add $1$ and $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{2} i i}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $i$ to the power of $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{2} (i^{1} i)}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Raise $i$ to the power of $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{2} (i^{1} i^{1})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{2} i^{1 + 1}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Add $1$ and $1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {\sqrt{3}}^{2} i^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + {(3^{\frac{1}{2}})}^{2} i^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + 3^{\frac{1}{2} \cdot 2} i^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Combine $\frac{1}{2}$ and $2$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + 3^{\frac{2}{2}} i^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Cancel the common factor of $2$ .

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Cancel the common factor.

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + 3^{\frac{2}{2}} i^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + 3^{1} i^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Evaluate the exponent.

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + 3 i^{2}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite $i^{2}$ as $- 1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} + 3 \cdot - 1}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $3$ by $- 1$ .

$y = 1 - 1 \frac{1 - i \sqrt{3} - i \sqrt{3} - 3}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Subtract $3$ from $1$ .

$y = 1 - 1 \frac{- i \sqrt{3} - i \sqrt{3} - 2}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Subtract $i \sqrt{3}$ from $- i \sqrt{3}$ .

$y = 1 - 1 \frac{- 2 i \sqrt{3} - 2}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Reorder $- 2 i \sqrt{3}$ and $- 2$ .

$y = 1 - 1 \frac{- 2 - 2 i \sqrt{3}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Cancel the common factor of $- 2 - 2 i \sqrt{3}$ and $2$ .

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Factor $2$ out of $- 2$ .

$y = 1 - 1 \frac{2 \cdot - 1 - 2 i \sqrt{3}}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Factor $2$ out of $- 2 i \sqrt{3}$ .

$y = 1 - 1 \frac{2 \cdot - 1 + 2 (- i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Factor $2$ out of $2 (- 1) + 2 (- i \sqrt{3})$ .

$y = 1 - 1 \frac{2 (- 1 - i \sqrt{3})}{2} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Cancel the common factors.

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Factor $2$ out of $2$ .

$y = 1 - 1 \frac{2 (- 1 - i \sqrt{3})}{2 (1)} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Cancel the common factor.

$y = 1 - 1 \frac{2 (- 1 - i \sqrt{3})}{2 \cdot 1} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = 1 - 1 \frac{- 1 - i \sqrt{3}}{1} - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Divide $- 1 - i \sqrt{3}$ by $1$ .

$y = 1 - 1 (- 1 - i \sqrt{3}) - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Apply the distributive property.

$y = 1 - 1 \cdot - 1 - 1 (- i \sqrt{3}) - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $- 1$ .

$y = 1 + 1 - 1 (- i \sqrt{3}) - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 1 (- i \sqrt{3})$ .

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Multiply $- 1$ by $- 1$ .

$y = 1 + 1 + 1 (i \sqrt{3}) - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $\sqrt{3}$ by $1$ .

$y = 1 + 1 + \sqrt{3} i - 5 (- \frac{1 - i \sqrt{3}}{2}) - 9$

Multiply $- 5 (- \frac{1 - i \sqrt{3}}{2})$ .

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Multiply $- 1$ by $- 5$ .

$y = 1 + 1 + \sqrt{3} i + 5 \frac{1 - i \sqrt{3}}{2} - 9$

Combine $5$ and $\frac{1 - i \sqrt{3}}{2}$ .

$y = 1 + 1 + \sqrt{3} i + \frac{5 (1 - i \sqrt{3})}{2} - 9$

Add $1$ and $1$ .

$y = 2 + \sqrt{3} i + \frac{5 (1 - i \sqrt{3})}{2} - 9$

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = 2 \cdot \frac{2}{2} + \frac{5 (1 - i \sqrt{3})}{2} + \sqrt{3} i - 9$

Combine $2$ and $\frac{2}{2}$ .

$y = \frac{2 \cdot 2}{2} + \frac{5 (1 - i \sqrt{3})}{2} + \sqrt{3} i - 9$

Combine the numerators over the common denominator.

$y = \frac{9 - 5 i \sqrt{3}}{2} + \sqrt{3} i - 9$

To write $\sqrt{3} i$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \frac{9 - 5 i \sqrt{3}}{2} + \sqrt{3} i \cdot \frac{2}{2} - 9$

Combine $\sqrt{3} i$ and $\frac{2}{2}$ .

$y = \frac{9 - 5 i \sqrt{3}}{2} + \frac{\sqrt{3} i \cdot 2}{2} - 9$

Combine the numerators over the common denominator.

$y = \frac{9 - 3 i \sqrt{3}}{2} - 9$

To write $- 9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \frac{9 - 3 i \sqrt{3}}{2} - 9 \cdot \frac{2}{2}$

Combine $- 9$ and $\frac{2}{2}$ .

$y = \frac{9 - 3 i \sqrt{3}}{2} + \frac{- 9 \cdot 2}{2}$

Combine the numerators over the common denominator.

$y = \frac{- 9 - 3 i \sqrt{3}}{2}$

Rewrite $- 9$ as $- 1 (9)$ .

$y = \frac{- 1 (9) - 3 i \sqrt{3}}{2}$

Factor $- 1$ out of $- 3 i \sqrt{3}$ .

$y = \frac{- 1 (9) - (3 i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (9) - (3 i \sqrt{3})$ .

$y = \frac{- 1 (9 + 3 i \sqrt{3})}{2}$

Move the negative in front of the fraction.

$y = - \frac{9 + 3 i \sqrt{3}}{2}$

Evaluate $y$ when $x = - \frac{1 + i \sqrt{3}}{2}$ .

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Substitute $- \frac{1 + i \sqrt{3}}{2}$ for $x$ .

$y = {(- \frac{1 + i \sqrt{3}}{2})}^{3} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Simplify ${(- \frac{1 + i \sqrt{3}}{2})}^{3} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$ .

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Simplify each term.

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Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1 + i \sqrt{3}}{2}$ .

$y = {(- 1)}^{3} {(\frac{1 + i \sqrt{3}}{2})}^{3} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the product rule to $\frac{1 + i \sqrt{3}}{2}$ .

$y = {(- 1)}^{3} \frac{{(1 + i \sqrt{3})}^{3}}{2^{3}} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $- 1$ to the power of $3$ .

$y = - \frac{{(1 + i \sqrt{3})}^{3}}{2^{3}} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $2$ to the power of $3$ .

$y = - \frac{{(1 + i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Use the Binomial Theorem.

$y = - \frac{1^{3} + 3 \cdot 1^{2} (i \sqrt{3}) + 3 \cdot 1 {(i \sqrt{3})}^{2} + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Simplify each term.

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One to any power is one.

$y = - \frac{1 + 3 \cdot 1^{2} (i \sqrt{3}) + 3 \cdot 1 {(i \sqrt{3})}^{2} + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

One to any power is one.

$y = - \frac{1 + 3 \cdot 1 (i \sqrt{3}) + 3 \cdot 1 {(i \sqrt{3})}^{2} + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $3$ by $1$ .

$y = - \frac{1 + 3 (i \sqrt{3}) + 3 \cdot 1 {(i \sqrt{3})}^{2} + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $3$ by $1$ .

$y = - \frac{1 + 3 i \sqrt{3} + 3 {(i \sqrt{3})}^{2} + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the product rule to $i \sqrt{3}$ .

$y = - \frac{1 + 3 i \sqrt{3} + 3 (i^{2} {\sqrt{3}}^{2}) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite $i^{2}$ as $- 1$ .

$y = - \frac{1 + 3 i \sqrt{3} + 3 (- 1 {\sqrt{3}}^{2}) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$y = - \frac{1 + 3 i \sqrt{3} + 3 (- 1 {(3^{\frac{1}{2}})}^{2}) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = - \frac{1 + 3 i \sqrt{3} + 3 (- 1 \cdot 3^{\frac{1}{2} \cdot 2}) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Combine $\frac{1}{2}$ and $2$ .

$y = - \frac{1 + 3 i \sqrt{3} + 3 (- 1 \cdot 3^{\frac{2}{2}}) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Cancel the common factor of $2$ .

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Cancel the common factor.

$y = - \frac{1 + 3 i \sqrt{3} + 3 (- 1 \cdot 3^{\frac{2}{2}}) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = - \frac{1 + 3 i \sqrt{3} + 3 (- 1 \cdot 3^{1}) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Evaluate the exponent.

$y = - \frac{1 + 3 i \sqrt{3} + 3 (- 1 \cdot 3) + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $3$ .

$y = - \frac{1 + 3 i \sqrt{3} + 3 \cdot - 3 + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $3$ by $- 3$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 + {(i \sqrt{3})}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the product rule to $i \sqrt{3}$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 + i^{3} {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Factor out $i^{2}$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 + i^{2} \cdot i {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite $i^{2}$ as $- 1$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 - 1 \cdot i {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite $- 1 i$ as $- i$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 - i {\sqrt{3}}^{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite ${\sqrt{3}}^{3}$ as ${(3^{3})}^{\frac{1}{2}}$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 - i \sqrt{3^{3}}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $3$ to the power of $3$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 - i \sqrt{27}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite $27$ as $3^{2} \cdot 3$ .

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Factor $9$ out of $27$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 - i \sqrt{9 (3)}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite $9$ as $3^{2}$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 - i \sqrt{3^{2} \cdot 3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Pull terms out from under the radical.

$y = - \frac{1 + 3 i \sqrt{3} - 9 - i (3 \sqrt{3})}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $3$ by $- 1$ .

$y = - \frac{1 + 3 i \sqrt{3} - 9 - 3 i \sqrt{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Subtract $9$ from $1$ .

$y = - \frac{3 i \sqrt{3} - 8 - 3 i \sqrt{3}}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Subtract $3 i \sqrt{3}$ from $3 i \sqrt{3}$ .

$y = - \frac{0 - 8}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Subtract $8$ from $0$ .

$y = - \frac{- 8}{8} - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Divide $- 8$ by $8$ .

$y = - - 1 - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $- 1$ .

$y = 1 - 2 {(- \frac{1 + i \sqrt{3}}{2})}^{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1 + i \sqrt{3}}{2}$ .

$y = 1 - 2 ({(- 1)}^{2} {(\frac{1 + i \sqrt{3}}{2})}^{2}) - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the product rule to $\frac{1 + i \sqrt{3}}{2}$ .

$y = 1 - 2 ({(- 1)}^{2} \frac{{(1 + i \sqrt{3})}^{2}}{2^{2}}) - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $- 1$ to the power of $2$ .

$y = 1 - 2 (1 \frac{{(1 + i \sqrt{3})}^{2}}{2^{2}}) - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $\frac{{(1 + i \sqrt{3})}^{2}}{2^{2}}$ by $1$ .

$y = 1 - 2 \frac{{(1 + i \sqrt{3})}^{2}}{2^{2}} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $2$ to the power of $2$ .

$y = 1 - 2 \frac{{(1 + i \sqrt{3})}^{2}}{4} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Cancel the common factor of $2$ .

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Factor $2$ out of $- 2$ .

$y = 1 + 2 (- 1) \frac{{(1 + i \sqrt{3})}^{2}}{4} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Factor $2$ out of $4$ .

$y = 1 + 2 \cdot - 1 \frac{{(1 + i \sqrt{3})}^{2}}{2 \cdot 2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Cancel the common factor.

$y = 1 + 2 \cdot - 1 \frac{{(1 + i \sqrt{3})}^{2}}{2 \cdot 2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = 1 - 1 \frac{{(1 + i \sqrt{3})}^{2}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite ${(1 + i \sqrt{3})}^{2}$ as $(1 + i \sqrt{3}) (1 + i \sqrt{3})$ .

$y = 1 - 1 \frac{(1 + i \sqrt{3}) (1 + i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Expand $(1 + i \sqrt{3}) (1 + i \sqrt{3})$ using the FOIL Method.

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Apply the distributive property.

$y = 1 - 1 \frac{1 (1 + i \sqrt{3}) + i \sqrt{3} (1 + i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the distributive property.

$y = 1 - 1 \frac{1 \cdot 1 + 1 (i \sqrt{3}) + i \sqrt{3} (1 + i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the distributive property.

$y = 1 - 1 \frac{1 \cdot 1 + 1 (i \sqrt{3}) + i \sqrt{3} \cdot 1 + i \sqrt{3} (i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Simplify and combine like terms.

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Simplify each term.

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Multiply $1$ by $1$ .

$y = 1 - 1 \frac{1 + 1 (i \sqrt{3}) + i \sqrt{3} \cdot 1 + i \sqrt{3} (i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $i \sqrt{3}$ by $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} \cdot 1 + i \sqrt{3} (i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $i$ by $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i \sqrt{3} (i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $i \sqrt{3} (i \sqrt{3})$ .

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Raise $i$ to the power of $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{1} i \sqrt{3} \sqrt{3}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $i$ to the power of $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{1} i^{1} \sqrt{3} \sqrt{3}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{1 + 1} \sqrt{3} \sqrt{3}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Add $1$ and $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{2} \sqrt{3} \sqrt{3}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $\sqrt{3}$ to the power of $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{2} ({\sqrt{3}}^{1} \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Raise $\sqrt{3}$ to the power of $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{2} ({\sqrt{3}}^{1} {\sqrt{3}}^{1})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{2} {\sqrt{3}}^{1 + 1}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Add $1$ and $1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} + i^{2} {\sqrt{3}}^{2}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite $i^{2}$ as $- 1$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 1 {\sqrt{3}}^{2}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 1 {(3^{\frac{1}{2}})}^{2}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 1 \cdot 3^{\frac{1}{2} \cdot 2}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Combine $\frac{1}{2}$ and $2$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 1 \cdot 3^{\frac{2}{2}}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Cancel the common factor of $2$ .

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Cancel the common factor.

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 1 \cdot 3^{\frac{2}{2}}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 1 \cdot 3^{1}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Evaluate the exponent.

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 1 \cdot 3}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $3$ .

$y = 1 - 1 \frac{1 + i \sqrt{3} + i \sqrt{3} - 3}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Subtract $3$ from $1$ .

$y = 1 - 1 \frac{i \sqrt{3} + i \sqrt{3} - 2}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Add $i \sqrt{3}$ and $i \sqrt{3}$ .

$y = 1 - 1 \frac{2 i \sqrt{3} - 2}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Reorder $2 i \sqrt{3}$ and $- 2$ .

$y = 1 - 1 \frac{- 2 + 2 i \sqrt{3}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Cancel the common factor of $- 2 + 2 i \sqrt{3}$ and $2$ .

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Factor $2$ out of $- 2$ .

$y = 1 - 1 \frac{2 \cdot - 1 + 2 i \sqrt{3}}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Factor $2$ out of $2 i \sqrt{3}$ .

$y = 1 - 1 \frac{2 \cdot - 1 + 2 (i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Factor $2$ out of $2 (- 1) + 2 (i \sqrt{3})$ .

$y = 1 - 1 \frac{2 (- 1 + i \sqrt{3})}{2} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Cancel the common factors.

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Factor $2$ out of $2$ .

$y = 1 - 1 \frac{2 (- 1 + i \sqrt{3})}{2 (1)} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Cancel the common factor.

$y = 1 - 1 \frac{2 (- 1 + i \sqrt{3})}{2 \cdot 1} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite the expression.

$y = 1 - 1 \frac{- 1 + i \sqrt{3}}{1} - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Divide $- 1 + i \sqrt{3}$ by $1$ .

$y = 1 - 1 (- 1 + i \sqrt{3}) - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Apply the distributive property.

$y = 1 - 1 \cdot - 1 - 1 (i \sqrt{3}) - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $- 1$ by $- 1$ .

$y = 1 + 1 - 1 (i \sqrt{3}) - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Rewrite $- 1 (i \sqrt{3})$ as $- (i \sqrt{3})$ .

$y = 1 + 1 - (i \sqrt{3}) - 5 (- \frac{1 + i \sqrt{3}}{2}) - 9$

Multiply $- 5 (- \frac{1 + i \sqrt{3}}{2})$ .

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Multiply $- 1$ by $- 5$ .

$y = 1 + 1 - i \sqrt{3} + 5 \frac{1 + i \sqrt{3}}{2} - 9$

Combine $5$ and $\frac{1 + i \sqrt{3}}{2}$ .

$y = 1 + 1 - i \sqrt{3} + \frac{5 (1 + i \sqrt{3})}{2} - 9$

Add $1$ and $1$ .

$y = 2 - i \sqrt{3} + \frac{5 (1 + i \sqrt{3})}{2} - 9$

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = - i \sqrt{3} + 2 \cdot \frac{2}{2} + \frac{5 (1 + i \sqrt{3})}{2} - 9$

Combine $2$ and $\frac{2}{2}$ .

$y = - i \sqrt{3} + \frac{2 \cdot 2}{2} + \frac{5 (1 + i \sqrt{3})}{2} - 9$

Combine the numerators over the common denominator.

$y = - i \sqrt{3} + \frac{9 + 5 i \sqrt{3}}{2} - 9$

To write $- i \sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = - i \sqrt{3} \cdot \frac{2}{2} + \frac{9 + 5 i \sqrt{3}}{2} - 9$

Combine $- i \sqrt{3}$ and $\frac{2}{2}$ .

$y = \frac{- i \sqrt{3} \cdot 2}{2} + \frac{9 + 5 i \sqrt{3}}{2} - 9$

Combine the numerators over the common denominator.

$y = \frac{9 + 3 i \sqrt{3}}{2} - 9$

To write $- 9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \frac{9 + 3 i \sqrt{3}}{2} - 9 \cdot \frac{2}{2}$

Combine $- 9$ and $\frac{2}{2}$ .

$y = \frac{9 + 3 i \sqrt{3}}{2} + \frac{- 9 \cdot 2}{2}$

Combine the numerators over the common denominator.

$y = \frac{- 9 + 3 i \sqrt{3}}{2}$

Rewrite $- 9$ as $- 1 (9)$ .

$y = \frac{- 1 (9) + 3 i \sqrt{3}}{2}$

Factor $- 1$ out of $3 i \sqrt{3}$ .

$y = \frac{- 1 (9) - (- 3 i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (9) - (- 3 i \sqrt{3})$ .

$y = \frac{- 1 (9 - 3 i \sqrt{3})}{2}$

Move the negative in front of the fraction.

$y = - \frac{9 - 3 i \sqrt{3}}{2}$

List all of the solutions.

$y = 3, x = 4$

$y = - \frac{9 + 3 i \sqrt{3}}{2}, x = - \frac{1 - i \sqrt{3}}{2}$

$y = - \frac{9 - 3 i \sqrt{3}}{2}, x = - \frac{1 + i \sqrt{3}}{2}$