Why is $x^{p/q}$ ill-defined for $x<0$. 
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 $x=-1$.
 A: $$y=x^{p/q}$$ is understood as a solution of
$$y^q=x^p,$$ which makes sense when the sign are compatible. For negative $x$, this excludes odd $p$ with even $q$.
Ambiguity arises when the fraction $p/q$ can be simplified by $2$, like
$$\frac{2p}{2q}=\frac pq,$$ and though the first fraction is even/even (giving a positive power), the second can be odd/even (undefined) or even/odd (negative).
A better definition could be to enforce simplification of the fraction,
$$y^{q/\gcd(p,q)}=x^{p/gcd(p,q)}$$ which avoids the ambiguity.

All cases can be summarized by


*

*$(-1)^{1/1}\to -1$

*$(-1)^{1/2}\to \text{undefined}$

*$(-1)^{2/1}\to +1$

*$(-1)^{2/2}\to(-1)^{1/1}=-1$

Without this convention, the "natural" rule
$$x^{p/q}=\sqrt[q]{x^p}={\sqrt[q]x}^p$$ doesn't hold.
A: There are some well-known ambiguities in defining say $(-1)^{1/2}$ because there are two candidates, $i$ and $-i$, just as there are two candidates for $2^{1/2}$ and by convention we choose the positive one.  However, we can adopt a convention whereby the candidate with positive imaginary part is chosen, and so define $(-1)^{1/2}=i$.  From there we can go to $(-1)^{n/2}$ for integers $n$.
It's harder to extend the convention further, but for even $m>2$, we can say that $(-1)^{1/m}=\exp(\pi i/m)$.  If $m$ is odd, we can adopt the same convention, but it's kind of counterintuitive, since $(-1)^{1/m}$ will not be $-1$ as we expect, for example $(-1)^{1/3}$ will be $\exp(\pi i/3)$ and not $-1$.  Alternatively, we can say that $(-1)^{1/m}=-1$ for all odd $m$.  I think it is the difficulty making these conventions coexist that is the source of the difficulties you see.
When you write $(-1)^{1/3} \neq (-1)^{2/6}$, I assume you mean to use a definition like $(-1)^{2/6}=((-1)^{1/6})^2$.  With my $\exp$ convention, I think you do get consistency, i.e., $\exp(\pi i/6)^2$ $=\exp(\pi i/3)$ $=(-1)^{1/3}$. 
