Answer
The solutions for $x$ are:
$2~(cos~20^{\circ}+i~sin~20^{\circ})$
$2~(cos~140^{\circ}+i~sin~140^{\circ})$
$2~(cos~260^{\circ}+i~sin~260^{\circ})$
Work Step by Step
$x^3-(4+4i~\sqrt{3}) = 0$
$x^3 = (4+4i~\sqrt{3})$
$x^3 = 8~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})$
$x = [8~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})]^{1/3}$
Let $z = 8~(\frac{1}{2}+\frac{i~\sqrt{3}}{2})$
$z = 8~(cos~60^{\circ}+i~sin~60^{\circ})$
$r = 8$ and $\theta = 60^{\circ}$
We can use this equation to find the cube roots:
$z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
When k = 0:
$z^{1/3} = 8^{1/3}~[cos(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$
$z^{1/3} = 2~(cos~20^{\circ}+i~sin~20^{\circ})$
When k = 1:
$z^{1/3} = 8^{1/3}~[cos(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$
$z^{1/3} = 2~(cos~140^{\circ}+i~sin~140^{\circ})$
When k = 2:
$z^{1/3} = 8^{1/3}~[cos(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{60^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$
$z^{1/3} = 2~(cos~260^{\circ}+i~sin~260^{\circ})$
