Algebra Examples

$(b^{2}) = b^{8}$

Step 1

Subtract $b^{8}$ from both sides of the equation.

$b^{2} - b^{8} = 0$

Step 2

Factor the left side of the equation.

Tap for more steps...

Step 2.1

Factor $b^{2}$ out of $b^{2} - b^{8}$ .

Tap for more steps...

Step 2.1.1

Multiply by $1$ .

$b^{2} \cdot 1 - b^{8} = 0$

Step 2.1.2

Factor $b^{2}$ out of $- b^{8}$ .

$b^{2} \cdot 1 + b^{2} (- b^{6}) = 0$

Step 2.1.3

Factor $b^{2}$ out of $b^{2} \cdot 1 + b^{2} (- b^{6})$ .

$b^{2} (1 - b^{6}) = 0$

Step 2.2

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

$b^{2} (1^{3} - b^{6}) = 0$

Step 2.3

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

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

Step 2.4

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 $a = 1$ and $b = b^{2}$ .

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

Step 2.5

Factor.

Tap for more steps...

Step 2.5.1

Simplify.

Tap for more steps...

Step 2.5.1.1

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

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

Step 2.5.1.2

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

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

Step 2.5.1.3

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

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

Step 2.5.2

Remove unnecessary parentheses.

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

Step 2.6

One to any power is one.

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

Step 2.7

Multiply the exponents in ${(b^{2})}^{2}$ .

Tap for more steps...

Step 2.7.1

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

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

Step 2.7.2

Multiply $2$ by $2$ .

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

Step 2.8

Factor.

Tap for more steps...

Step 2.8.1

Rewrite $1 + b^{2} + b^{4}$ in a factored form.

Tap for more steps...

Step 2.8.1.1

Rewrite the middle term.

$b^{2} (1 + b) (1 - b) (1 + 2 \cdot (1 b^{2}) - b^{2} + b^{4}) = 0$

Step 2.8.1.2

Rearrange terms.

$b^{2} (1 + b) (1 - b) (1 + 2 \cdot (1 b^{2}) + b^{4} - b^{2}) = 0$

Step 2.8.1.3

Factor first three terms by perfect square rule.

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

Step 2.8.1.4

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

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

Step 2.8.1.5

Simplify.

Tap for more steps...

Step 2.8.1.5.1

Reorder terms.

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

Step 2.8.1.5.2

Reorder terms.

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

Step 2.8.2

Remove unnecessary parentheses.

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

Step 3

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

$b^{2} = 0$

$1 + b = 0$

$1 - b = 0$

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

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

Step 4

Set $b^{2}$ equal to $0$ and solve for $b$ .

Tap for more steps...

Step 4.1

Set $b^{2}$ equal to $0$ .

$b^{2} = 0$

Step 4.2

Solve $b^{2} = 0$ for $b$ .

Tap for more steps...

Step 4.2.1

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

$b = \pm \sqrt{0}$

Step 4.2.2

Simplify $\pm \sqrt{0}$ .

Tap for more steps...

Step 4.2.2.1

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

$b = \pm \sqrt{0^{2}}$

Step 4.2.2.2

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

$b = \pm 0$

Step 4.2.2.3

Plus or minus $0$ is $0$ .

$b = 0$

Step 5

Set $1 + b$ equal to $0$ and solve for $b$ .

Tap for more steps...

Step 5.1

Set $1 + b$ equal to $0$ .

$1 + b = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

$b = - 1$

Step 6

Set $1 - b$ equal to $0$ and solve for $b$ .

Tap for more steps...

Step 6.1

Set $1 - b$ equal to $0$ .

$1 - b = 0$

Step 6.2

Solve $1 - b = 0$ for $b$ .

Tap for more steps...

Step 6.2.1

Subtract $1$ from both sides of the equation.

$- b = - 1$

Step 6.2.2

Divide each term in $- b = - 1$ by $- 1$ and simplify.

Tap for more steps...

Step 6.2.2.1

Divide each term in $- b = - 1$ by $- 1$ .

$\frac{- b}{- 1} = \frac{- 1}{- 1}$

Step 6.2.2.2

Simplify the left side.

Tap for more steps...

Step 6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{b}{1} = \frac{- 1}{- 1}$

Step 6.2.2.2.2

Divide $b$ by $1$ .

$b = \frac{- 1}{- 1}$

Step 6.2.2.3

Simplify the right side.

Tap for more steps...

Step 6.2.2.3.1

Divide $- 1$ by $- 1$ .

$b = 1$

Step 7

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

Tap for more steps...

Step 7.1

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

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

Step 7.2

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

Tap for more steps...

Step 7.2.1

Use the quadratic formula to find the solutions.

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

Step 7.2.2

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

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

Step 7.2.3

Simplify.

Tap for more steps...

Step 7.2.3.1

Simplify the numerator.

Tap for more steps...

Step 7.2.3.1.1

One to any power is one.

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

Step 7.2.3.1.2

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

Tap for more steps...

Step 7.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 7.2.3.1.3

Subtract $4$ from $1$ .

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

Step 7.2.3.1.4

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

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

Step 7.2.3.1.5

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

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

Step 7.2.3.1.6

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

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

Step 7.2.3.2

Multiply $2$ by $1$ .

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

Step 7.2.4

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

Tap for more steps...

Step 7.2.4.1

Simplify the numerator.

Tap for more steps...

Step 7.2.4.1.1

One to any power is one.

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

Step 7.2.4.1.2

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

Tap for more steps...

Step 7.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.2.4.1.2.2

Multiply $- 4$ by $1$ .

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

Step 7.2.4.1.3

Subtract $4$ from $1$ .

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

Step 7.2.4.1.4

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

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

Step 7.2.4.1.5

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

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

Step 7.2.4.1.6

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

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

Step 7.2.4.2

Multiply $2$ by $1$ .

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

Step 7.2.4.3

Change the $\pm$ to $+$ .

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

Step 7.2.4.4

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

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

Step 7.2.4.5

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

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

Step 7.2.4.6

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

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

Step 7.2.4.7

Move the negative in front of the fraction.

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

Step 7.2.5

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

Tap for more steps...

Step 7.2.5.1

Simplify the numerator.

Tap for more steps...

Step 7.2.5.1.1

One to any power is one.

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

Step 7.2.5.1.2

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

Tap for more steps...

Step 7.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.2.5.1.2.2

Multiply $- 4$ by $1$ .

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

Step 7.2.5.1.3

Subtract $4$ from $1$ .

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

Step 7.2.5.1.4

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

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

Step 7.2.5.1.5

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

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

Step 7.2.5.1.6

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

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

Step 7.2.5.2

Multiply $2$ by $1$ .

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

Step 7.2.5.3

Change the $\pm$ to $-$ .

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

Step 7.2.5.4

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

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

Step 7.2.5.5

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

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

Step 7.2.5.6

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

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

Step 7.2.5.7

Move the negative in front of the fraction.

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

Step 7.2.6

The final answer is the combination of both solutions.

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

Step 8

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

Tap for more steps...

Step 8.1

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

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

Step 8.2

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

Tap for more steps...

Step 8.2.1

Use the quadratic formula to find the solutions.

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

Step 8.2.2

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

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

Step 8.2.3

Simplify.

Tap for more steps...

Step 8.2.3.1

Simplify the numerator.

Tap for more steps...

Step 8.2.3.1.1

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

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

Step 8.2.3.1.2

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

Tap for more steps...

Step 8.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 8.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 8.2.3.1.3

Subtract $4$ from $1$ .

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

Step 8.2.3.1.4

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

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

Step 8.2.3.1.5

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

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

Step 8.2.3.1.6

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

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

Step 8.2.3.2

Multiply $2$ by $1$ .

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

Step 8.2.4

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

Tap for more steps...

Step 8.2.4.1

Simplify the numerator.

Tap for more steps...

Step 8.2.4.1.1

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

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

Step 8.2.4.1.2

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

Tap for more steps...

Step 8.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 8.2.4.1.2.2

Multiply $- 4$ by $1$ .

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

Step 8.2.4.1.3

Subtract $4$ from $1$ .

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

Step 8.2.4.1.4

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

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

Step 8.2.4.1.5

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

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

Step 8.2.4.1.6

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

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

Step 8.2.4.2

Multiply $2$ by $1$ .

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

Step 8.2.4.3

Change the $\pm$ to $+$ .

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

Step 8.2.5

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

Tap for more steps...

Step 8.2.5.1

Simplify the numerator.

Tap for more steps...

Step 8.2.5.1.1

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

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

Step 8.2.5.1.2

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

Tap for more steps...

Step 8.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 8.2.5.1.2.2

Multiply $- 4$ by $1$ .

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

Step 8.2.5.1.3

Subtract $4$ from $1$ .

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

Step 8.2.5.1.4

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

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

Step 8.2.5.1.5

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

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

Step 8.2.5.1.6

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

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

Step 8.2.5.2

Multiply $2$ by $1$ .

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

Step 8.2.5.3

Change the $\pm$ to $-$ .

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

Step 8.2.6

The final answer is the combination of both solutions.

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

Step 9

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

$b = 0, - 1, 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$