# How do you factor completely 8y^3 + 27?

Mar 25, 2018

$\left(2 y + 3\right) \left(4 {y}^{2} - 6 y + 9\right)$

#### Explanation:

Sum of cubes: ${a}^{3} + {b}^{3}$

Factored sum of cubes: $\left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

In this case, a is $2 y$ and b is $3$

${\left(2 y\right)}^{3} + {3}^{3} = 8 {y}^{3} + 27$

Plug 2y and 3 in for a and b, respectively

$\left(2 y + 3\right) \left({\left(2 y\right)}^{2} - 2 y \cdot 3 + {3}^{2}\right)$

$\left(2 y + 3\right) \left(4 {y}^{2} - 6 y + 9\right)$