Answer
The other two cube roots are $2-2i\sqrt{3}$ and $-4$
Work Step by Step
We can find $(2+2i\sqrt{3})^2$:
$(2+2i\sqrt{3})^2 = 4-12+8i\sqrt{3}$
$(2+2i\sqrt{3})^2 = -8+8i\sqrt{3}$
We can find $(2+2i\sqrt{3})^3$:
$(2+2i\sqrt{3})^3 = (2+2i\sqrt{3})(-8+8i\sqrt{3})$
$(2+2i\sqrt{3})^3 = -16-48-16i\sqrt{3}+16i\sqrt{3}$
$(2+2i\sqrt{3})^3 = -64$
Let $z = -64 = 64~(-1+0~i)$
$z = 64~(cos~180^{\circ}+i~sin~180^{\circ})$
$r = 64$ and $\theta = 180^{\circ}$
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:
$z^{1/3} = 64^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$
$z^{1/3} = 4~(cos~60^{\circ}+i~sin~60^{\circ})$
$z^{1/3} = 2+2i\sqrt{3}$
When k = 1:
$z^{1/3} = 64^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$
$z^{1/3} = 4~(cos~180^{\circ}+i~sin~180^{\circ})$
$z^{1/3} = -4$
When k = 2:
$z^{1/3} = 64^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$
$z^{1/3} = 4~(cos~300^{\circ}+i~sin~300^{\circ})$
$z^{1/3} = 2-2i\sqrt{3}$
