Solve for x 9x-x^3=0

$9 x - x^{3} = 0$

Step 1

Factor the left side of the equation.

Factor $x$ out of $9 x - x^{3}$ .

Factor $x$ out of $9 x$ .

$x \cdot 9 - x^{3} = 0$

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

$x \cdot 9 + x (- x^{2}) = 0$

Factor $x$ out of $x \cdot 9 + x (- x^{2})$ .

$x (9 - x^{2}) = 0$

Rewrite $9$ as $3^{2}$ .

$x (3^{2} - x^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = x$ .

$x ((3 + x) (3 - x)) = 0$

Remove unnecessary parentheses.

$x (3 + x) (3 - x) = 0$

Step 2

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

$x = 0$

$3 + x = 0$

$3 - x = 0$

Step 3

Set $x$ equal to $0$ .

$x = 0$

Step 4

Set $3 + x$ equal to $0$ and solve for $x$ .

Set $3 + x$ equal to $0$ .

$3 + x = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Set $3 - x$ equal to $0$ and solve for $x$ .

Set $3 - x$ equal to $0$ .

$3 - x = 0$

Solve $3 - x = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

Step 6

The final solution is all the values that make $x (3 + x) (3 - x) = 0$ true.

$x = 0, - 3, 3$

Solve for x 9x-x^3=0

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