Why do I get an imaginary result for the cube root of a negative number? I have a function that includes the phrase $(-x)^{1/3}$. It seems like this should always evaluate to $-(x^{1/3})$. For example, $-1 \cdot -1 \cdot -1 = -1$, so it seems that $(-1)^{1/3}$ should equal $-1$.
When I plug $(-1)^{(1/3)}$ into something like Mathematica, I get: 
0.5 + 0.866025i

Cubing this answer does in fact compute to $-1$.
Is this a situation, like $\sqrt4$, where there are two valid answers , $\{-2, 2\}$?
 A: The end result is that $x^y$ is technically a multi-valued function whenever $y$ is not an integer.  This is often overlooked in favor of considering only the principal root (smallest argument greater than or equal to zero) as is the case with the square root (which will result in real numbers only in the case that $x$ is a non-negative real number).  In the further case that $y$ is an irrational number, then there are infinitely many solutions.
Here, I will attempt to go into more detail about the case where $y$ is a simple fraction with $1$ as the numerator.

De Moivre's Theorem for fractional powers: for a complex number in polar form with $r$ a non-negative real number, one has
$$(r(cos\theta+i\sin\theta))^{\frac{1}{n}}=\sqrt[n]{r}\left[\cos\left(\frac{2k\pi+\theta}{n}\right)+i\sin\left(\frac{2k\pi+\theta}{n}\right)\right]~~\text{for}~k=0,1,2,\dots,n-1$$

Graphically, if one were to want to find the results of $z^{\frac{1}{n}}$ for some complex number $z$, plot it on the complex plane.  Take the magnitude of $z$
and replace it with the magnitude of $\sqrt[n]{|z|}$.  Take the angle of $z$ and replace it with $\frac{1}{n}$ of that angle.  From there, draw a regular $n$-gon with one of the points being the just found point.  All of the corners will be the solutions.

Pictured above is the original number $z$ (in red), and its three cube roots (in blue).
A: Use $e^{i\pi}=-1$. Then $(e^{i\pi})^{1/3}=e^{i\pi/3}=-1^{1/3}$
De Moivre's gives $e^{i\theta}=\cos(\theta)+i\sin(\theta)$
If $\theta=\frac{\pi}{3}$, then it follows that
$(-1)^{1/3}=e^{i(\pi/3)}=\frac12+\frac{i\sqrt3}{2}\approx0.5 + 0.866025i$
