What does $(-1)^{2/3}$ equal? I figured that according to the exponent laws it should equal $1$ since: $$(-1)^{2/3} = ((-1)^2)^{1/3} = 1^{1/3} = 1$$
 But according to wolframalpha and google it equals something imaginary. Why is this?
 A: This is so because you can not use rules of exponents for negative bases.
A similar argument as of yours will lead to following $$i=(-1)^{1/2}=(-1)^{2/4}=1^{1/4}=1$$
A: In general, fractional powers are multivalued functions.  For  complex numbers $a$ and $r$, $a^r$ is defined as $\exp(r \log(a))$, where $\log(a)$ is any branch of the natural logarithm of $a$.  In this case, $\log(-1) = i \pi + 2 n i \pi$ for arbitrary integer $n$, so 
$$(-1)^{2/3} = \exp((2 i \pi + 4 n i \pi)/3)$$
has three possible values: $\exp(0) = 1$, $\exp(2 i\pi/3) = \frac{-1+i \sqrt{3}}{2}$, $\exp(4 i \pi/3) = \frac{-1 - i \sqrt{3}}{2}$.
I suspect Wolfram Alpha is using the principal branch $n=0$.
A: Have you ever seen the equality $e^{\pi i} = -1$? We can actually use this to solve this puzzle!
Let's try  $(-1)^{\frac{2}{3}} = (e^{\pi i})^{\frac{2}{3}} = e^{\frac{2}{3}\pi i}$. If you cube that, you do get $1 = e^{2 \pi i}$.
For more info, you can learn about roots of unity , complex roots, and the Euler Identity.
A: If we are only looking for real solutions, we can allow $(-1)^{\frac ab}$ for $\frac ab$ written in reduced form and $b$ odd.
$$(-1)^{\frac ab}=\begin{cases}1 &\text{ for } a\text{ even } \\
-1 &\text{ for } a\text{ odd } \end{cases}$$
This way, we will never have a fraction with even on top and bottom, which causes problems. If we allow $b$ to be even, then we can end up taking an even root of a negative number which isn't allowed in the "real world". 
Robert Israel gives the full answer and the reason why Wolfram Alpha gives the answer that it does.
