What is the principal cubic root of $-8$? According to my book it should be a real number, and according to WolframAlpha it should be $1+1.73i$

What is the correct answer?
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.
Consider $x^3 =  - 8 $. It &quot;seems&quot; we can just say $x = -2$ because $(- 2)^3 = -8$ but $- 2$ is not the official Primary Solution!
Using De Moivre's theorem again:
$$x^3 =  - 8$$
$$( r \operatorname{cis}(\theta) )^3 = -8$$
$$r^3 \operatorname{cis}(3\theta) = 8\operatorname{cis}( 180^\circ + 360^\circ n)$$
$$r^3  = 8 \text{ and } 3\theta = 180^\circ + 360^\circ n $$
$$r = 2 \text{ and } \theta = 60^\circ + 120^\circ n = 60^\circ, 180^\circ, 300^\circ $$
$$x_1  =  2\operatorname{cis}( 60^\circ ) = 1 + i\sqrt3$$
$$x_2  = 2\operatorname{cis}(180^\circ) =  - 2$$
$$x_3  = 2\operatorname{cis}(240^\circ) =  1 - i\sqrt3$$
The Primary Solution is $x_1 = 1 + i\sqrt3   ≈ 1 + 1.732i$
So the cube root of $-8$ or $(-8)^{\frac13}=1+1.732i$ and NOT $-2$.