# How do you factor completely  x^3 - 9x ?

Aug 1, 2016

${x}^{3} - 9 x = x \left(x + 3\right) \left(x - 3\right)$

#### Explanation:

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

= $x \left({x}^{2} - 9\right)$

= $x \left({x}^{2} - {3}^{2}\right)$

Now using identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, we get

$x \left({x}^{2} - {3}^{2}\right)$

= $x \left(x + 3\right) \left(x - 3\right)$