# How do you factor 64x^3-8?

Feb 20, 2017

$8 \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$

#### Explanation:

Factor $64 {x}^{3} - 8$

First factor out the GCF $8$.

$8 \left(8 {x}^{3} - 1\right)$

Rewrite the expression as a difference of cubes, where $a = 2 x$ and $b = 1$.

$8 \left({\left(2 x\right)}^{3} - {1}^{3}\right)$

Use the difference of cubes to answer this question:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

8((2x)^3-1^3)=8(2x-1)((2x^2+(2x*1)+1^2)=

$8 \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right) =$

$8 \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right)$