# How do you factor 8000 + y^3?

Jan 1, 2016

Use the sum of cubes identity to find:

$8000 + {y}^{3} = \left(20 + y\right) \left(400 - 20 y + {y}^{2}\right)$

#### Explanation:

Both $8000 = {20}^{3}$ and ${y}^{3}$ are perfect cubes.

Use the sum of cubes identity:

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

with $a = 20$ and $b = y$ as follows:

$8000 + {y}^{3}$

$= {20}^{3} + {y}^{3}$

$= \left(20 + y\right) \left({20}^{2} - 20 y + {y}^{2}\right)$

$= \left(20 + y\right) \left(400 - 20 y + {y}^{2}\right)$