# How do you factor x^6-4?

May 16, 2017

${x}^{6} - 4 = \left({x}^{3} + 2\right) \left({x}^{3} - 2\right)$

#### Explanation:

Both terms are perfect squares and they are being subtracted.

We have the difference of squares which is factored as :

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

${x}^{6} - 4 = \left({x}^{3} + 2\right) \left({x}^{3} - 2\right)$

Note that $6$ is not a square, but a base with an even index is a square.

$\sqrt{{x}^{6}} = {x}^{3}$