# How do you factor 27x³ - 64y³?

Jan 31, 2016

$\left(3 x - 4 y\right) \left(9 {x}^{2} + 12 x y + 16 {y}^{2}\right)$

#### Explanation:

This is a difference of 2 cubes which can be factorised as

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

for $27 {x}^{3} - 64 {y}^{3} , a = 3 x \textcolor{b l a c k}{\text{ and }} b = 4 y$

substituting these into the above expression to obtain

$27 {x}^{3} - 64 {y}^{3} = \left(3 x - 4 y\right) \left(9 {x}^{2} + 12 x y + 16 {y}^{2}\right)$