# How do you factor 64x^3 - y^6?

Jun 15, 2018

$64 {x}^{3} - {y}^{6} = \left(4 x - {y}^{2}\right) \left(16 {x}^{2} + 4 x {y}^{2} + {y}^{4}\right)$

#### Explanation:

As $64 {x}^{3} - {y}^{6}$ can be written as ${\left(4 x\right)}^{3} - {\left({y}^{2}\right)}^{3}$, we can use the identity

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

Hence $64 {x}^{3} - {y}^{6} = {\left(4 x\right)}^{3} - {\left({y}^{2}\right)}^{3}$

= (4x-y^2)((4x)^2+4x×y^2+(y^2)^2)

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