# How do you find all the real and complex roots of x^3 + 8 = 0?

Jan 19, 2016

$x = - 2 , 1 \pm \sqrt{3} i$

#### Explanation:

Split this apart using the sum of cubes identity.

$\left(x + 2\right) \left({x}^{2} - 2 x + 4\right) = 0$

Now, you know that either

$x + 2 = 0 \textcolor{w h i t e}{\times \times} \text{or} \textcolor{w h i t e}{\times \times} {x}^{2} - 2 x + 4 = 0$

Solving the first of these gives that $x = - 2$.

The other two can be found through applying the quadratic formula on the quadratic:

$x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm 2 \sqrt{3} i}{2} = 1 \pm \sqrt{3} i$