Calculus Examples

$x^{6} - 2 x^{3} + 1$

Step 1

Set $x^{6} - 2 x^{3} + 1$ equal to $0$ .

$x^{6} - 2 x^{3} + 1 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Rewrite $x^{6}$ as ${(x^{3})}^{2}$ .

${(x^{3})}^{2} - 2 x^{3} + 1 = 0$

Step 2.1.2

Let $u = x^{3}$ . Substitute $u$ for all occurrences of $x^{3}$ .

$u^{2} - 2 u + 1 = 0$

Step 2.1.3

Factor using the perfect square rule.

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Step 2.1.3.1

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

$u^{2} - 2 u + 1^{2} = 0$

Step 2.1.3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 2.1.3.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 1 + 1^{2} = 0$

Step 2.1.3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 1$ .

${(u - 1)}^{2} = 0$

Step 2.1.4

Replace all occurrences of $u$ with $x^{3}$ .

${(x^{3} - 1)}^{2} = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Rewrite $1$ as $1^{3}$ .

${(x^{3} - 1^{3})}^{2} = 0$

Step 2.2.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Step 2.2.3

Simplify.

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Step 2.2.3.1

Multiply $x$ by $1$ .

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

Step 2.2.3.2

One to any power is one.

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

Step 2.2.4

Apply the product rule to $(x - 1) (x^{2} + x + 1)$ .

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

Step 2.3

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

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

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

Step 2.4

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set ${(x - 1)}^{2}$ equal to $0$ .

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

Step 2.4.2

Solve ${(x - 1)}^{2} = 0$ for $x$ .

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Step 2.4.2.1

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.4.2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

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

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Step 2.5.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x^{2} + x + 1 = \pm \sqrt{0}$

Step 2.5.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.5.2.2.1

Rewrite $0$ as $0^{2}$ .

$x^{2} + x + 1 = \pm \sqrt{0^{2}}$

Step 2.5.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

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

Step 2.5.2.2.3

Plus or minus $0$ is $0$ .

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

Step 2.5.2.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.5.2.4

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

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 2.5.2.5

Simplify.

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Step 2.5.2.5.1

Simplify the numerator.

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Step 2.5.2.5.1.1

One to any power is one.

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

Step 2.5.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.5.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.5.2.5.1.2.2

Multiply $- 4$ by $1$ .

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

Step 2.5.2.5.1.3

Subtract $4$ from $1$ .

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

Step 2.5.2.5.1.4

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

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

Step 2.5.2.5.1.5

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

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

Step 2.5.2.5.1.6

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

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

Step 2.5.2.5.2

Multiply $2$ by $1$ .

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

Step 2.5.2.6

The final answer is the combination of both solutions.

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

Step 2.6

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

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

Step 3