cubic root of negative numbers Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of $-8$ would be $-2$ since $(-2)^3 = -8$.

But when I checked Wolfram Alpha for $\sqrt[3]{-8}$, real it tells me it doesn't exist.

I came to trust Wolfram Alpha so I thought I'd ask you guys, to explain the sense of that to me.
-8 has three cube roots: $ -2 $, $1 + i \sqrt{ 3 } $ and $1 - i \sqrt{ 3 } $. So you can't answer the question "Is $ \sqrt[3]{-8} $ real" without specifying which of them you're talking about.

For some reason, WolframAlpha is only giving $1 + i \sqrt{ 3 } $ as an answer -- that looks like a bug in WolframAlpha to me.
Although the equation  $x^3+8 = 0$  has actually three roots (real $-2$ and two conjugated complex roots) still the third root of $-8$ does not exist. Nth roots are defined only for nonnegative real values. Please, consider the $$(-8)^{1/3} = (-8)^{2/6}$$  that gives either  $64^{1/6} = 2$  or  $(\sqrt{  -8})^2 $ = nonsense.   Let $(-8)^{1/3} = (-8)^{2/6}$ you never get $(-2)$ as the result. Therefore Nth root exists only of nonnegative real numbers.