# How do you factor x^3+343?

Apr 23, 2016

${x}^{3} + 343 = \left(x + 7\right) \left({x}^{2} - 7 x + 49\right)$

#### Explanation:

The sum of cubes identity can be written:

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

We can use this with $a = x$ and $b = 7$ as follows:

${x}^{3} + 343$

$= {x}^{3} + {7}^{3}$

$= \left(x + 7\right) \left({x}^{2} - 7 x + {7}^{2}\right)$

$= \left(x + 7\right) \left({x}^{2} - 7 x + 49\right)$

The remaining quadratic factor can only be factored further with Complex coefficients, which we can express in terms of the primitive Complex cube root of $1$:

$= \left(x + 7\right) \left(x + 7 \omega\right) \left(x + 7 {\omega}^{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$