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

May 11, 2016

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

#### Explanation:

Both ${x}^{3}$ and $64 = {4}^{3}$ are perfect squares.

So we can use the sum of cubes identity:

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

with $a = x$ and $b = 4$ as follows:

${x}^{3} + 64$

$= {x}^{3} + {4}^{3}$

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

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