I am confused if the function $f(x) = x^{4\over5}$ is an even function. If we only consider the real roots, then the function is even in the sense that $f(-x) = f(x)$.
However, since $x^{1\over 5}$ is likely to have some non-real roots for $x \in \mathbb{R}$, and hence it may be the case when $x^{1\over 5} \neq (-x)^{1\over 5}$, would this $f(x)$ still fit into the definition of an even function?
On
Split up the fraction. Either $$f(x)=(x^4)^\frac15$$ Where $(x^4)=(-x)^4$ so these are equivalent, Or: $$f(x)=(x^\frac15)^4$$ Where $x^\frac15=-(-x)^\frac15$ (take the fifth power on both sides to see this), but using $(x)^4=(-x)^4$ these are equivalent as well. It is understood that when selecting which root to take, you take the same root on both sides.
For example, if you were calculating values of $f(x) =x^\frac13+(-x)^\frac13$, to say that $f(8)=2+(-1+i\sqrt3)=1+i\sqrt3$ is totally wrong.
On
The function is even. It seems like what's really in question is $(-1)^{\frac{p}{q}}$ with $p$ even, $q$ odd. This equals $1$. To see this, consider that $(-1)^{\frac{1}{q}}=-1$ is the unique real solution to $x^q=-1$. Then $(-1)^{\frac{p}{q}}=((-1)^{\frac{1}{q}})^p=(-1)^p=1$ since $p$ is even.
This is also true if you do the order in reverse in that $(-1)^p=1$ and $1^\frac{1}{q}=1$ is the unique real solution to $x^q=1$.
In conclusion, $(-x)^\frac{p}{q}=x^\frac{p}{q}(-1)^\frac{p}{q}=x^\frac{p}{q}$.
When you define a function you need to define the domain. When we talk of even and odd functions we are normally talking of functions $\Bbb {R \to R}$. The behavior of the complex values (not roots) of $x^{1/5}$ does not matter. As real functions $x^{1/5}$ is odd and $x^{4/5}$ is even.