Solve by Factoring 8y^3-64=0

$8 y^{3} - 64 = 0$

Step 1

Factor $8$ out of $8 y^{3} - 64$ .

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

$8 (y^{3}) - 64 = 0$

Factor $8$ out of $- 64$ .

$8 y^{3} + 8 \cdot - 8 = 0$

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

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

Step 2

Rewrite $8$ as $2^{3}$ .

$8 (y^{3} - 2^{3}) = 0$

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 $b = 2$ .

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

Step 4

Factor.

Simplify.

Move $2$ to the left of $y$ .

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

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

$8 ((y - 2) (y^{2} + 2 y + 4)) = 0$

Remove unnecessary parentheses.

$8 (y - 2) (y^{2} + 2 y + 4) = 0$

Step 5

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

$y - 2 = 0$

$y^{2} + 2 y + 4 = 0$

Step 6

Set $y - 2$ equal to $0$ and solve for $y$ .

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

Add $2$ to both sides of the equation.

$y = 2$

Step 7

Set $y^{2} + 2 y + 4$ equal to $0$ and solve for $y$ .

Set $y^{2} + 2 y + 4$ equal to $0$ .

$y^{2} + 2 y + 4 = 0$

Solve $y^{2} + 2 y + 4 = 0$ for $y$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 2$ , and $c = 4$ into the quadratic formula and solve for $y$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

$y = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$y = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

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

$y = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$y = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

$y = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$y = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

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

$y = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$y = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

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

Change the $\pm$ to $+$ .

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

$y = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$y = \frac{- 2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{- 2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

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

$y = \frac{- 2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$y = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

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

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

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

Step 8

The final solution is all the values that make $8 (y - 2) (y^{2} + 2 y + 4) = 0$ true.

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

Solve by Factoring 8y^3-64=0

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