# How do you factor  2y^3 -128?

May 29, 2016

$2 {y}^{3} - 128 = 2 \left(y - 4\right) \left({y}^{2} + 4 y + 16\right)$

#### Explanation:

Separate out the common scalar factor $2$, then factor as a difference of cubes:

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

with $a = y$ and $b = 4$ as follows:

$2 {y}^{3} - 128$

$= 2 \left({y}^{3} - 64\right)$

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

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

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