# How do you factor x^3+8y^3?

Sep 23, 2015

Use the sum of cubes identity ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$ to find:

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

#### Explanation:

Use the sum of cubes identity with $a = x$ and $b = 2 y$:

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

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

= (x+2y(x^2-2xy+4y^2)