# How do you factor  27y^3+64?

May 2, 2016

$27 {y}^{3} + 64 = \left(3 y + 4\right) \left(9 {y}^{2} - 12 y + 16\right)$

#### Explanation:

Note that both $27 {y}^{3} = {\left(3 y\right)}^{3}$ and $64 = {4}^{3}$ are perfect cubes.

So it is natural to use the sum of cubes identity:

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

with $a = 3 y$ and $b = 4$ as follows:

$27 {y}^{3} + 64$

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

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

$= \left(3 y + 4\right) \left(9 {y}^{2} - 12 y + 16\right)$