# How do you factor y^3-64?

Jan 30, 2016

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

#### Explanation:

${y}^{3} - 64$ is a difference of cubes, a^3– b^3 = (a – b)(a^2 + ab + b^2), where $a = y$ and $b = 4$.

Rewrite the equation.

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

Simplify.

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