Answer
The three possible values for $x$ are:
$5~(cos~60^{\circ}+i~sin~60^{\circ})$
$5~(cos~180^{\circ}+i~sin~180^{\circ})$
$5~(cos~300^{\circ}+i~sin~300^{\circ})$
Work Step by Step
$x^3+125 = 0$
$x^3 = -125$
$x = (-125)^{1/3}$
$x = [125(-1+0~i)]^{1/3}$
Let $z = 125(-1 + 0~i)$
$z = 125(cos~180^{\circ}+i~sin~180^{\circ})$
$r = 125$ 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:
$z^{1/3} = 125^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$
$z^{1/3} = 5~(cos~60^{\circ}+i~sin~60^{\circ})$
When k = 1:
$z^{1/3} = 125^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$
$z^{1/3} = 5~(cos~180^{\circ}+i~sin~180^{\circ})$
When k = 2:
$z^{1/3} = 125^{1/3}~[cos(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{180^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$
$z^{1/3} = 5~(cos~300^{\circ}+i~sin~300^{\circ})$
