# How do you factor completely 64-x^3?

Sep 2, 2016

$64 - {x}^{3} = \left(4 - x\right) \left(16 + 4 x + {x}^{2}\right)$

#### Explanation:

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Note both $64 = {4}^{3}$ and ${x}^{3}$ are perfect cubes.

So we find:

$64 - {x}^{3} = {4}^{3} - {x}^{3}$

$\textcolor{w h i t e}{64 - {x}^{3}} = \left(4 - x\right) \left({4}^{2} + 4 x + {x}^{2}\right)$

$\textcolor{w h i t e}{64 - {x}^{3}} = \left(4 - x\right) \left(16 + 4 x + {x}^{2}\right)$