# How do you factor 2x^3+162 ?

Aug 3, 2015

$2 {x}^{3} + 162 = 2 \left({x}^{3} + {3}^{4}\right)$

$= 2 \left(x + {3}^{\frac{4}{3}}\right) \left({x}^{2} - {3}^{\frac{4}{3}} x + {3}^{\frac{8}{3}}\right)$

#### Explanation:

First separate out the common scalar factor $2$ to get:

$2 {x}^{3} + 162 = 2 \left({x}^{3} + 81\right) = 2 \left({x}^{3} + {3}^{4}\right)$

To factor this further, we want to treat $\left({x}^{3} + {3}^{4}\right)$ as a sum of cubes, then use the identity ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.

Unfortunately, $\sqrt[3]{{3}^{4}}$ is not rational. It can be expressed as $\sqrt[3]{81}$ or $3 \sqrt[3]{3}$ or ${3}^{\frac{4}{3}}$. I'll use the last of these.

Let $a = x$ and $b = {3}^{\frac{4}{3}}$ to find:

${x}^{3} + {3}^{4} = {a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

$= \left(x + {3}^{\frac{4}{3}}\right) \left({x}^{2} - {3}^{\frac{4}{3}} x + {3}^{\frac{8}{3}}\right)$

If we go beyond real numbers to complex numbers we can factor this further as:

${x}^{3} + {3}^{4} = \left(x + {3}^{\frac{4}{3}}\right) \left(x + {3}^{\frac{4}{3}} \omega\right) \left(x + {3}^{\frac{4}{3}} {\omega}^{2}\right)$

where $\omega = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ is the primitive complex cube root of $1$.