# How do you factor 1000x^3+27?

Aug 28, 2016

$1000 {x}^{3} + 27 = \left(10 x + 3\right) \left(100 {x}^{2} - 30 x + 9\right)$

#### 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 $a = 10 x$ and $b = 3$ as follows:

$1000 {x}^{3} + 27$

$= {\left(10 x\right)}^{3} + {3}^{3}$

$= \left(10 x + 3\right) \left({\left(10 x\right)}^{2} - \left(10 x\right) \left(3\right) + {3}^{2}\right)$

$= \left(10 x + 3\right) \left(100 {x}^{2} - 30 x + 9\right)$

This is as far as we can go with Real coefficients. If we allow Complex coefficients then it can be factored further as:

$= \left(10 x + 3\right) \left(10 x + 3 \omega\right) \left(10 x + 3 {\omega}^{2}\right)$

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