Find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$ Given that $x^n - \frac1{x^n}$ is expressible as a polynomial in $x - \frac1x$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$

My brain is not working with me today!  I don't know how to start this problem.  Solutions are greatly appreciated.

$$\left(x-\frac{1}{x}\right)^5 = x^5-5x^3+10x-\frac{10}{x}+\frac{5}{x^3}-\frac{1}{x^5} $$
$$\left(x-\frac{1}{x}\right)^3 = x^3-3x+\frac{3}{x}-\frac{1}{x^3}$$
hence
$$ \left(x-\frac{1}{x}\right)^5 + 5\left(x-\frac{1}{x}\right)^3 +5\left(x-\frac{1}{x}\right)=x^5-\frac{1}{x^5}$$
and
$$ P(z) = \color{red}{z^5+5z^3+5z}.$$
