# How do you get the complex cube root of 8?

Jan 3, 2016

The cube roots of $8$ are $2$, $2 \omega$ and $2 {\omega}^{2}$ where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

#### Explanation:

Here are the cube roots of $8$ plotted in the Complex plane on the circle of radius $2$:

graph{(x^2+y^2-4)((x-2)^2+y^2-0.01)((x+1)^2+(y-sqrt(3))^2-0.01)((x+1)^2+(y+sqrt(3))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}

They can be written as:

$2 \left(\cos \left(0\right) + i \sin \left(0\right)\right) = 2$

$2 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right) = - 1 + \sqrt{3} i = 2 \omega$

$2 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right) = - 1 - \sqrt{3} i = 2 {\omega}^{2}$

One way of finding these cube roots of $8$ is to find all of the roots of ${x}^{3} - 8 = 0$.

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$= \frac{- 2 \pm \sqrt{- 12}}{2}$

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