Answer
Every real number has exactly one real cube root.
Work Step by Step
Let $x$ be any real number.
Let $z = x+0~i$
$z = x~(1+0~i)$
$z = x~(cos~0^{\circ}+i~sin~0^{\circ})$
$r = x$ and $\theta = 0^{\circ}$
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:
$z^{1/3} = x^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(0)}{3})]$
$z^{1/3} = x^{1/3}~(cos~0^{\circ}+i~sin~0^{\circ})$
$z^{1/3} = x^{1/3}$
When k = 1:
$z^{1/3} = x^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(1)}{3})]$
$z^{1/3} = x^{1/3}~(cos~120^{\circ}+i~sin~120^{\circ})$
When k = 2:
$z^{1/3} = x^{1/3}~[cos(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})+i~sin(\frac{0^{\circ}}{3}+\frac{(360^{\circ})(2)}{3})]$
$z^{1/3} = x^{1/3}~(cos~240^{\circ}+i~sin~240^{\circ})$
The three cube roots of $x$ are:
$x^{1/3}$
$x^{1/3}~(cos~120^{\circ}+i~sin~120^{\circ})$
$x^{1/3}~(cos~240^{\circ}+i~sin~240^{\circ})$
Only $x^{1/3}$ is a real root. The other two roots have a non-zero imaginary part. Therefore, every real number has exactly one real cube root.
