What is the domain of the function $f(x)=\sqrt[3]{x^3-x}$? Let $f$ be: $f(x) = \sqrt[3]{x^3 -x}$, an exercise book asked for the domain of definition. Isn't it over $\mathbb R$. The book solution stated $Df = [-1,0] \cup [1, +\infty[$ 
I don t get it. Can you explain? 
 A: If your book reaches the domain $[-1,0]\cup[1,+\infty)$, it must be because the book only considers $\sqrt[3]{\phantom{X}}$ to be defined when the argument is a non-negative real.
Books (and people) differ in how they consider $\sqrt[N]{\phantom X}$ to be defined.
Some people find it okay to define odd roots on the entire real line -- after all, $x\mapsto x^N$ is a bijection on $\mathbb R$ when $N$ is positive odd, and every such bijection has a perfectly fine inverse.
Other people prefer to restrict these functions to non-negative reals, no matter what $N$ is -- partially to avoid creating a (confusing?) distinction between odd and even $N$, partially for more subtle reasons that unfortunately are not apparent when one first learns about roots.
(For even subtler reasons, one might even want to reserve the root notation to arguments that are strictly positive, such that $\sqrt 0$ is considered undefined. It is somewhat rare to take that position consistently, though).
You'll just have to live with the fact that such questions cannot be answered without knowing which convention for the root sign is to be used. (Arguably it is bad form to let a find-the-domain-of-this-expression exercise depend on such choices, but that's purely the textbook's fault, of course).
