Domain of $y=x^2\ (2 - x)^{2/3}$ and $y=x^2\ \sqrt[3]{(2 - x)^{2}}$. I have some problems with Domain of the following real function:
$$y=x^2\ (2 - x)^{2/3}$$
Since $t^{2/3}=\sqrt[3]{t^2}$, we can rewrite function as
$$y=x^2\ \sqrt[3]{(2 - x)^{2}}$$
thus Domain is $D=\mathbb R$. But, by Wolfram:
$$D=\{x \in\mathbb R : x\leq 2\}$$
Why? Maybe is $t^{2/3}=\sqrt[3]{t^2}$ not always possible?
Thanks in advance.
 A: For the set of complex numbers, the answer from Wolfram alpha is correct
But for the set of real numbers, your answer is correct.
A: By WolframAlpha $$\sqrt[3]{x^2}\not\equiv x^{2/3}$$
It has to do with complex numbers. For $x<0$ values of these two functions differ, as for complex numbers the rule $$(a^b)^c=a^{bc}$$ is not always true. It comes from the fact that complex roots are multivalued.
Domain of your function depends on how you define $x^{m/n}$. In school, teachers define it as $$x^{m/n}=\sqrt[n]{x^m}$$
This is pretty intuitive definition but, unfortunately, not consistent for negative $x$. When dealing with complex numbers, this expression is multivalued and its principal value is defined differently, as 
$$x^{m/n}=|x|^{m/n}(\cos(\tfrac{m}n\cdot \operatorname{Arg}(x))+i\sin(\tfrac{m}n\cdot\operatorname{Arg}(x))=|x|^{m/n}\exp(\tfrac{m}n\cdot i\operatorname{Arg}(x))$$
When dealing with first definition, your domain is correct, it is $$d=\mathbb R$$
whereas when dealing with the second definition, for $x<0$ the expression gives complex numbers, so domain is $$D:x\geqslant 0$$
A: As per kamil09875's answer, if you're dealing with complex numbers the two functions are not equivalent. If you're dealing with real numbers and live in a world where $\sqrt[3]{-1} = -1$, then both functions are defined over all $\mathbb{R}$ and equivalent.  See this discussion on $(a^b)^c = a^{bc}$.
