Solve for b (b^2)=b^8

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

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.

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

Multiply by $1$ .

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

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

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

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

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

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

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

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

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

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 $b = b^{2}$ .

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

Factor.

Simplify.

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

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

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 $b = b$ .

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

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

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

Remove unnecessary parentheses.

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

One to any power is one.

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

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

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

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

Multiply $2$ by $2$ .

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

Factor.

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

Rewrite the middle term.

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

Rearrange terms.

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

Factor first three terms by perfect square rule.

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

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 $b = b$ .

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

Simplify.

Reorder terms.

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

Reorder terms.

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

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 $b$ .

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

$b^{2} = 0$

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

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

$b = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

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

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

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

$b = \pm 0$

Plus or minus $0$ is $0$ .

$b = 0$

Step 5

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

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

$1 + b = 0$

Subtract $1$ from both sides of the equation.

$b = - 1$

Step 6

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

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

$1 - b = 0$

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

Subtract $1$ from both sides of the equation.

$- b = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $b$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$b = 1$

Step 7

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

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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 $b$ .

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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.

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

Solve for b (b^2)=b^8

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