Trigonometry (11th Edition) Clone

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