From which set does the number $\sqrt[3]{-1}$ belong to? I was trying to draw the function $f(x)=\sqrt[3]{x^2(6-x)}$ by hand (I'm in my first year of engineering; having Calculus I; this drawing is actually an exercise given for my class) and used WolframAlpha to see if I've got it right. Well, I was expecting to miss a few things on my first try, but I didn't even consider the possibility that I would start getting its domain wrong (I thought it was all the real numbers, however, the program said it was all real numbers equal to or below $6$). So, trying to understand what I'd missed, I've concluded that Wolfram must see $\sqrt[3]{-1}$ as only as a complex number when I swore it was at least real (since the equation $x^3 = -1$ has at least a real number that solves it, i.e. $-1$). I mean, if Wolfram is right (and I'm supposing it is), then $f(x=7)$ does not return the "real number" $\sqrt[3]{-49}$ (I don't even know if this number is real anymore, and I'm getting more and more confused whilst writing this text), and my professor has put a wrong graph as an answer for the exercise above. Look, although I wanted the ability to understand what I'm not understanding so that I could be, at least, didactic, I'll try to resume my confusion with this question (and with it, I'll try to understand on my own why Wolfram didn't consider numbers like $x=7$ for $f(x)$ domain): what in heavens is this $\sqrt[3]{-1}$?
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PS: sorry if I've said anything wrong... I don't speak English fluently.
 A: When working in the real numbers, $\sqrt[3]{x}$ denotes the unique number $a$ such that $a^3=x$. Since $(-1)^3=-1$, we have $\sqrt[3]{-1}=-1$.
When working in the complex numbers, the notation $\sqrt[3]{z}$ is imprecise. Every complex number has three cube roots, and there is no natural way of defining the 'principal' cube root of a complex number. It is sensible, however, to define $\sqrt[3]{z}$ in the complex numbers if $z$ also happens to be a real number. Then, $\sqrt[3]{z}$ can be defined in the way above.
That being said, sometimes we define the 'principal' $n$-th root of a complex number in the following way. If $z=re^{i\theta}$, where $\theta$ is the principal argument of $z$*, and $r$ is the magnitude of $z$, then
$$
\sqrt[n]{z} = \sqrt[n]{r} \cdot e^{i\theta/n} \, ,
$$
where $\sqrt[n]{r}$ denotes the principal real root of $r$. This means that $z \mapsto \sqrt[n]{z}$ is not a continuous function, but we still might choose to adopt the notation $\sqrt[n]{z}$ as a matter of convenience. This is probably the root that Wolfram Alpha was referring to:
$$
\sqrt[3]{-1} = \sqrt[3]{1e^{i\pi}} = \sqrt[3]{1} \cdot e^{(i\pi)/3} = e^{(i\pi)/3} \approx 0.5 + 0.866i \, .
$$

*Again, the 'principal' argument of a complex number involves a branch cut, where we require that $\theta \in (-\pi,\pi]$. Again, this does not define a continuous function.
A: Wolfram Alpha is slightly eccentric in this regard. The cube root function is a perfectly respectable function from $\Bbb R$ to $\Bbb R$, but the default behaviour of Wolfram Alpha is to evaluate cube roots of negative reals as complex numbers.
However, you can override this behaviour, by clicking on "Use the real-valued root instead". Then you can have Wolfram Alpha plot the function for any range you like.
A: You are completely right. $x^3=a$, with $a<0$ has a real root, so the problem must lay in the way you are writing it in Mathematica. Try using the function CubeRoot, and see if it works. If not,  I would advise you to share your code in the question.
