# How do you factor 8y^3-125?

Feb 27, 2017

$8 {y}^{3} - 125 = \left(2 y - 5\right) \left(4 {y}^{2} + 10 y + 25\right)$

#### Explanation:

The difference of cubes identity can be written:

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

Use this with $a = 2 y$ and $b = 5$ as follows:

$8 {y}^{3} - 125 = {\left(2 y\right)}^{3} - {5}^{3}$

$\textcolor{w h i t e}{8 {y}^{3} - 125} = \left(2 y - 5\right) \left({\left(2 y\right)}^{2} + \left(2 y\right) \left(5\right) + {5}^{2}\right)$

$\textcolor{w h i t e}{8 {y}^{3} - 125} = \left(2 y - 5\right) \left(4 {y}^{2} + 10 y + 25\right)$