Solve by Factoring y^3=16y

$y^{3} = 16 y$

Step 1

Subtract $16 y$ from both sides of the equation.

$y^{3} - 16 y = 0$

Step 2

Factor $y$ out of $y^{3} - 16 y$ .

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

$y \cdot y^{2} - 16 y = 0$

Factor $y$ out of $- 16 y$ .

$y \cdot y^{2} + y \cdot - 16 = 0$

Factor $y$ out of $y \cdot y^{2} + y \cdot - 16$ .

$y (y^{2} - 16) = 0$

Step 3

Rewrite $16$ as $4^{2}$ .

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

Step 4

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 4$ .

$y ((y + 4) (y - 4)) = 0$

Remove unnecessary parentheses.

$y (y + 4) (y - 4) = 0$

Step 5

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

$y = 0$

$y + 4 = 0$

$y - 4 = 0$

Step 6

Set $y$ equal to $0$ .

$y = 0$

Step 7

Set $y + 4$ equal to $0$ and solve for $y$ .

Set $y + 4$ equal to $0$ .

$y + 4 = 0$

Subtract $4$ from both sides of the equation.

$y = - 4$

Step 8

Set $y - 4$ equal to $0$ and solve for $y$ .

Set $y - 4$ equal to $0$ .

$y - 4 = 0$

Add $4$ to both sides of the equation.

$y = 4$

Step 9

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

$y = 0, - 4, 4$