# How do you factor 4x^3 + 36?

May 9, 2016

$4 {x}^{3} + 36 = 4 \left(x + \sqrt[3]{9}\right) \left({x}^{2} - \sqrt[3]{9} x + 3 \sqrt[3]{3}\right)$

#### Explanation:

First separate out the common scalar factor $4$:

$4 {x}^{3} + 36 = 4 \left({x}^{3} + 9\right)$

Note that ${x}^{3}$ is a perfect cube, but $9$ is not - at least not a cube of a rational number.

We can still use the sum of cubes identity:

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

with $a = x$ and $b = \sqrt[3]{9}$ as follows:

${x}^{3} + 9$

$= {x}^{3} + {\left(\sqrt[3]{9}\right)}^{3}$

$= \left(x + \sqrt[3]{9}\right) \left({x}^{2} - x \left(\sqrt[3]{9}\right) + {\left(\sqrt[3]{9}\right)}^{2}\right)$

$= \left(x + \sqrt[3]{9}\right) \left({x}^{2} - \sqrt[3]{9} x + \sqrt[3]{81}\right)$

$= \left(x + \sqrt[3]{9}\right) \left({x}^{2} - \sqrt[3]{9} x + 3 \sqrt[3]{3}\right)$

So:

$4 {x}^{3} + 36 = 4 \left(x + \sqrt[3]{9}\right) \left({x}^{2} - \sqrt[3]{9} x + 3 \sqrt[3]{3}\right)$