# What are the cube roots of 8?

Mar 4, 2018

Solve ${x}^{3} - 8 = 0$

#### Explanation:

Use either known factoring ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$ or divide ${x}^{3} - 8$ by $x - 2$ to find the quedratic factor.

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

${x}^{3} - 8 = 0$ if and only if

$x - 2 = 0$ or ${x}^{2} + 2 x + 4 = 0$

So one solution is $2$

The quadratic may be solved by completing the square of by formula

$x = \frac{- \left(2\right) \pm \sqrt{{\left(2\right)}^{2} - 4 \left(1\right) \left(4\right)}}{2 \left(1\right)}$

$= - 1 \pm i \sqrt{3}$