# How do you factor x^6 - 8y^3?

Jan 10, 2016

${x}^{6} - 8 {y}^{3} = \left({x}^{2} - 2 y\right) \left({x}^{4} + 2 {x}^{2} y + 4 {y}^{2}\right)$

#### Explanation:

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

Therefore by setting $a = {x}^{2}$ and $b = 2 y$:

${x}^{6} - 8 {y}^{3} = \left({x}^{2} - 2 y\right) \left({x}^{4} + 2 {x}^{2} y + 4 {y}^{2}\right)$

Don't think it does but it possibly could be factorised more.