Solve by Factoring 8y^3-8=0

$8 y^{3} - 8 = 0$

Step 1

Factor $8$ out of $8 y^{3} - 8$ .

Factor $8$ out of $8 y^{3}$ .

$8 (y^{3}) - 8 = 0$

Factor $8$ out of $- 8$ .

$8 (y^{3}) + 8 (- 1) = 0$

Factor $8$ out of $8 (y^{3}) + 8 (- 1)$ .

$8 (y^{3} - 1) = 0$

Step 2

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

$8 (y^{3} - 1^{3}) = 0$

Step 3

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 $a = y$ and $b = 1$ .

$8 ((y - 1) (y^{2} + y \cdot 1 + 1^{2})) = 0$

Step 4

Factor.

Simplify.

Multiply $y$ by $1$ .

$8 ((y - 1) (y^{2} + y + 1^{2})) = 0$

One to any power is one.

$8 ((y - 1) (y^{2} + y + 1)) = 0$

Remove unnecessary parentheses.

$8 (y - 1) (y^{2} + y + 1) = 0$

Step 5

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

$y - 1 = 0$

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

Step 6

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

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

$y - 1 = 0$

Add $1$ to both sides of the equation.

$y = 1$

Step 7

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

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

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

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

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 $y$ .

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

Simplify.

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $+$ .

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

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

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $-$ .

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

The final answer is the combination of both solutions.

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

Step 8

The final solution is all the values that make $8 (y - 1) (y^{2} + y + 1) = 0$ true.

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

Solve by Factoring 8y^3-8=0

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