# How do you factor 2w^3 + 54?

Dec 19, 2015

$2 \left(w + 3\right) \left({w}^{2} - 3 w + 9\right)$

#### Explanation:

First, factor out a common $2$.

$= 2 \left({w}^{3} + 27\right)$

Notice that $\left({w}^{3} + 27\right)$ is a sum of cubes, which follows the rule

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

Thus,

$2 \left({w}^{3} + 27\right) = 2 \left({\left(w\right)}^{3} + {\left(3\right)}^{3}\right) = 2 \left(w + 3\right) \left({w}^{2} - 3 w + 9\right)$