Is $a^{p/q}$ equal to $a^{2p/2q}$? Do we need to simplify $p/q$ to its lowest terms? I need a strict mathematical definition which proves one or another statement. For example:
$(-8)^{1/3} = -2$
Is $(-8)^{2/6}$ equal to $\sqrt[6]{(-8)^2} = 2$?
Or $(\sqrt[6]{-8})^2$, which is undefined?
Or $(-8)^{1/3} =-2$?
On
As long as $a\ge 0$, the function $a^x$ is well defined and therefore it doesn't matter how you represent x, either as $\frac 1 3$ or $\frac 2 6$, etc.
Once a hits negative numbers, you need to be careful because of the inability to take even powered roots of negative numbers, so for instance $(x^2)^\frac 1 2$ is not equal to x, but instead |x|.
$1/3$ and $2/6$ are the same number, so if we need to evaluate $(-8)^{1/3}$ and $(-8)^{2/6}$, we must make sure that the result is the same in each case. So yes, you need to reduce a fractional exponent to its lowest terms before applying it to a negative operand.
Having said that, the exponential function $x^y$ doesn't seem to be useful for negative $x$ and rational $y$, except perhaps when $y$ is of a special form to begin with, like $1/n$ for instance. So the question doesn't usually arise. Where did this question come from?