Is it possible to find a closed form for $A(x, p)=\frac{z}{x-z}+\frac{z^{p-1}}{x-z^{p-1}}$?

Question

Here $z=e^{2\pi i/p}$, where $p$ is prime, $x\in \mathbb{R}$ and $i^2=-1$. I know that the sum is real,but I can't find it's value. For example, if $x=2$ and $p=5$ we have the following:
$$A(2, 5)=\frac{z}{2-z}+\frac{z^{4}}{2-z^{4}}=\frac{1}{31}(3\sqrt{5}-13).$$
I would like to see a result of the form $A(x, p)=\frac{x-1}{x^p-1}\cdot (a\cos2\pi/p+b)$ but I am unable to prove it. I have managed to express $A(x, p)$ as an infinite series, as a sum of the form $2[1+x\cos m+x^2\cos2m+x^3\cos3m+\ldots+x^{p-1}\cos (p-1)m]$ and as another finite sum, but I am stuck.

Answer 1

$z^p=1$, so the result is $$\frac{z}{x-z}+\frac{z^{-1}}{x-z^{-1}}=\frac{2(x\cos\frac{2\pi}{p}-1)}{x^2-2x\cos\frac{2\pi}{p}+1}.$$