Solve by Factoring 7x^3-63x=0

$7 x^{3} - 63 x = 0$

Step 1

Factor $7 x$ out of $7 x^{3} - 63 x$ .

Factor $7 x$ out of $7 x^{3}$ .

$7 x (x^{2}) - 63 x = 0$

Factor $7 x$ out of $- 63 x$ .

$7 x (x^{2}) + 7 x (- 9) = 0$

Factor $7 x$ out of $7 x (x^{2}) + 7 x (- 9)$ .

$7 x (x^{2} - 9) = 0$

Step 2

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

$7 x (x^{2} - 3^{2}) = 0$

Step 3

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 $b = 3$ .

$7 x ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$7 x (x + 3) (x - 3) = 0$

Step 4

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

$x = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 5

Set $x$ equal to $0$ .

$x = 0$

Step 6

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 8

The final solution is all the values that make $7 x (x + 3) (x - 3) = 0$ true.

$x = 0, - 3, 3$