How to evaluate $(-2\sqrt2)^{2/3}$?

What is the exact value of this expression? $$\left ( -2 \sqrt 2 \right )^{2 \over 3}$$ Isn't $2$ one of the answer? Wolfram gets $-1+i \sqrt 3$. Is root multivariable function for complex number?

-
I think there are three roots. Wolfram Alpha may not give you all three roots though. To make sure Wolfram Alpha give you $2$, use parentheses like this: ((-2 * sqrt(2))^2)^(1/3) – Tunococ Aug 23 '12 at 2:47
Sure, $n$-th root of anything but $0$ is $n$-valued. – André Nicolas Aug 23 '12 at 2:49

Let $x = (-2 \sqrt 2)^{2 \over 3}$.

$x^3 = (-2\sqrt{2})^2 = 8$.

Let $\omega \neq 1$ be a root of $x^3 = 1$. Then, the roots of $x^3 = 8$ are $2, 2\omega, 2\omega^2$. Let's compute $\omega$.

$x^3 - 1 = (x - 1)(x^2 + x + 1)$.

Hence $\omega = \frac{-1 + i\sqrt{3}}{2}$ or $\frac{-1 - i\sqrt{3}}{2}$.

Hence the roots of $x^3 = 8$ are $2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$.

-
if all those values are result of expression, then is $2 = -1 + i \sqrt 3$. I mean what is the exact value of that expression? – Monkey D. Luffy Aug 23 '12 at 4:59
There are 3 exact values. – Makoto Kato Aug 23 '12 at 5:31
@user10254, obviously $\,2=-1+i\sqrt 3\,$ is false, as complex numbers are equal iff their real and imaginary are equal, corresp. As the answer shows, there are 3 "exact" values for that expression, and without any further conditions none is "more correct" or "better" than other. – DonAntonio Aug 23 '12 at 5:54
@DonAntonio how can an expression have three values? is it multivariable function? can you refer some topic for me to study? – Monkey D. Luffy Aug 23 '12 at 10:33
@user10254 , it's like the expression $\,4^{1/2}\,$: what is it? Certainly mathematicians have agreed to take the positive root, but this is mainly an agreement to make $\,f(x)=\sqrt x\,$ an actual function. When checking the possible values the above expression has we must state both $\,-2\,,\,2\,$ , and none of these values is more accurate or better than the other one. – DonAntonio Aug 23 '12 at 17:02