# How do you factor  64x^4 + xy^3?

Dec 22, 2016

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

#### Explanation:

The sum 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 $a = 4 x$ and $b = y$ as follows:

$64 {x}^{4} + x {y}^{3} = x \left(64 {x}^{3} + {y}^{3}\right)$

$\textcolor{w h i t e}{64 {x}^{4} + x {y}^{3}} = x \left({\left(4 x\right)}^{3} + {y}^{3}\right)$

$\textcolor{w h i t e}{64 {x}^{4} + x {y}^{3}} = x \left(4 x + y\right) \left({\left(4 x\right)}^{2} - \left(4 x\right) y + {y}^{2}\right)$

$\textcolor{w h i t e}{64 {x}^{4} + x {y}^{3}} = x \left(4 x + y\right) \left(16 {x}^{2} - 4 x y + {y}^{2}\right)$