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

May 3, 2018

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

#### Explanation:

This is an example of factoring perfect cubes.

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

$64$ is a perfect cube $4 \cdot 4 \cdot 4$
${x}^{3}$ is a perfect cube $x \cdot x \cdot x$
${y}^{3}$ is a perfect cube $y \cdot y \cdot y$

The factor pattern for factoring perfect cubes is always

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

The signs follow the acronym SOAP

First sign is the SAME as the expression
Second sign is the OPPOSITE of the expression
Third sign is ALWAYS POSITIVE

For this problem
$a = 4 x$ the root of $64 {x}^{3}$
$b = y$ the root of ${y}^{3}$

So plugging in we get

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

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