What is the principal cubic root of $-8$?

The cubic roots of $-8$ are: $$ 2e^{i\pi/3}=2(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}) \qquad 2e^{i\pi}=-2 \qquad 2e^{i2\pi/3}=2(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}) $$

it seems that WolframAlpha consider as principal root the root with the minimum value of the argument, i.e $e^{i\pi/3}$, but usually, if there is a real root this is considered the principal root.

Mathematicians (as opposed to elementary textbook writers) consider complex numbers. To make the principal value of the cube root continuous in as large a region of the complex plane as possible, and to have it agree with the conventional cube root for positive numbers, they choose the cube root with argument in $(-\pi/3 , \pi/3]$. This fits together with a systematic choice of principal value for logarithm, arctangent, and others.

There are three complex cubic roots of $-8$.

Namely $2 e^{i \pi/3}$, $2 e^{i (\pi+2 \pi) /3}$, $2 e^{i (\pi+4 \pi) /3} $. You get this using the polar form(s) of $-8$, that is $8 e^{i \pi}$ and more generally $8 e^{i (\pi+ 2k\pi)}$ for $k $ and integer.

The middle one is $-2$ and the first one is what Wolfram Alpha gives.

The rational for W|A making that choice is that it considers as principal root the one where the principal argument of $-8$, that is $\pi$, is divided by $3$.

The rational of your book should be that they want to be cubic roots of reals to be real.

It is a matter of definition/convention. In some sense I prefer the former, but would use the latter in the right context too.

In any case, you should follow your book or talk to your instructor about it, as otherwise there will be confusion.