algebra precalculus - Why is $x^{p/q}$ ill-defined for $x

This is probably a duplicate but I can't find, if you do let me know and I will delete.

Why is $x^{p/q}$ ill-defined for $x<0$.

I can see that it is, $(-1)^{1/3} \neq (-1)^{2/6}$, but why?

I define $x^{p/q}=\sqrt[q]{x^p}=(\sqrt[q]{x})^p$.

Also how does this affect calculus, in examples what about,

$$\int_{-1}^{0} \sqrt[3]{x} dx ~~~~?= ~~~~\int_{-1}^{0} x^{1/3} dx$$

$\frac{d}{dx}(\sqrt[3]{x})$ at $x=-1$.