Solve by Factoring x^3=36x

$x^{3} = 36 x$

Step 1

Subtract $36 x$ from both sides of the equation.

$x^{3} - 36 x = 0$

Step 2

Factor $x$ out of $x^{3} - 36 x$ .

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

$x \cdot x^{2} - 36 x = 0$

Factor $x$ out of $- 36 x$ .

$x \cdot x^{2} + x \cdot - 36 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 36$ .

$x (x^{2} - 36) = 0$

Step 3

Rewrite $36$ as $6^{2}$ .

$x (x^{2} - 6^{2}) = 0$

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 $a = x$ and $b = 6$ .

$x ((x + 6) (x - 6)) = 0$

Remove unnecessary parentheses.

$x (x + 6) (x - 6) = 0$

Step 5

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

$x = 0$

$x + 6 = 0$

$x - 6 = 0$

Step 6

Set $x$ equal to $0$ .

$x = 0$

Step 7

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 8

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 9

The final solution is all the values that make $x (x + 6) (x - 6) = 0$ true.

$x = 0, - 6, 6$