# How do you factor completely 1 + x^3?

Dec 20, 2015

Use the sum of cubes identity to find:

$1 + {x}^{3} = \left(1 + x\right) \left(1 - x + {x}^{2}\right)$

#### Explanation:

The sum of cubes identity may be written:

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

In our example, we have $a = 1$ and $b = x$ as follows:

$1 + {x}^{3}$

$= {1}^{3} + {x}^{3}$

$= \left(1 + x\right) \left({1}^{2} - \left(1\right) \left(x\right) + {x}^{2}\right)$

$= \left(1 + x\right) \left(1 - x + {x}^{2}\right)$

The remaining quadratic factor $\left(1 - x + {x}^{2}\right)$ cannot be factored into simpler factors with Real coefficients, but if you want a complete factorisation then you can do it with Complex coefficients:

$= \left(1 + x\right) \left(1 + \omega x\right) \left(1 + {\omega}^{2} x\right)$

or if you prefer:

$= \left(1 + x\right) \left(\omega + x\right) \left({\omega}^{2} + x\right)$

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