converting $\frac{-1 - \sqrt{3}i}{2^{2/3}}$ to $-(-2)^{1/3}$ I have a question. It might seem silly. 
Suppose I have $(x^3-2)$.
I know that $ 2^{1/3}$ is a root.  $\frac{-1 \pm \sqrt{3}i}{2^{2/3}}$ are also roots of $(x^3-2)$.
How do you convert $\frac{-1 - \sqrt{3}i}{2^{2/3}}$ to $-(-2)^{1/3}$?
Thanks.

The principal value of the third root of $-2\;$ is $$(-2)^{1/3}= |-2|^{1/3}\left(\cos(\frac{2}{3}\pi) + i \sin(\frac{2}{3}\pi)\right)
=\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Multiply with $-2$ to get
$$-(-2)^{1/3}=-\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Now multiply your expression with $2^{1/3}$ to get
$$\frac{-1 - \sqrt{3}i}{2^{2/3}}=-\frac{2^{1/3}}{2}\left(1 + \sqrt{3}i\right)$$
Therefore
$$\frac{-1 - \sqrt{3}i}{2^{2/3}}=-(-2)^{1/3}$$
