<s>[INST] 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?
 [/INST] 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 $2, -1 + i\sqrt{3}, -1 - i\sqrt{3}$.
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