# How do you factor 64a^3 - 8b^3?

May 30, 2015

This is a difference of cubes...

$64 {a}^{3} - 8 {b}^{3} = {\left(4 a\right)}^{3} - {\left(2 b\right)}^{3}$

Now ${A}^{3} - {B}^{3} = \left(A - B\right) \left({A}^{2} + A B + {B}^{2}\right)$

So

${\left(4 a\right)}^{3} - {\left(2 b\right)}^{3} = \left(\left(4 a\right) - \left(2 b\right)\right) \left({\left(4 a\right)}^{2} + \left(4 a\right) \left(2 b\right) + {\left(2 b\right)}^{2}\right)$

$= \left(4 a - 2 b\right) \left(16 {a}^{2} + 8 a b + 4 {b}^{2}\right)$

$= 8 \left(2 a - b\right) \left(4 {a}^{2} + 2 a b + {b}^{2}\right)$

The are no simpler factorings with real coefficients.