Find a Polynomial in $x-\frac1x$ Given that $x^n - (1/x^n)$ is expressible as a polynomial in $x - (1/x)$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P(x-(1/x)) = x^5 - (1/x)^5.$
To start, I factored, giving:
$P(x-\frac{1}{x}) = (x-\frac{1}{x})(x^4+\frac{1}{x^4}+x^2+\frac{1}{x^2}+1).$
However, I cannot find a way to connect this to $x-\frac{1}{x}$. What should I do?
 A: One has if we denote $Y_n=x^n-1/x^n$ and $Y_1=Y$
$$\begin{align}
\left(x-{1\over x}\right)^5
&=x^5-5x^3+10x-{10\over x}+{5\over x^3}-{1\over x^5}\\
&=Y_5-5Y_3+10Y
\end{align}$$
And
$$\begin{align}
\left(x-{1\over x}\right)^3
&=x^3-3x+{3\over x}-{1\over x^3}\\
&=Y_3-3Y
\end{align}$$
We then derive from the second identity that $5Y_3=5Y^3+15Y$ and therefore
$$\begin{align}x^5-{1\over x^5}&=Y_5\\&=Y^5+5Y_3-10Y\\&=Y^5+5Y^3+5Y\end{align}$$
A: Hint...write out the binomial expansions of $(x-\frac 1x)^5$ and $(x-\frac 1x)^3$ and rearrange terms
A: Try $P(x)=x^5+5x^3+5x $. The idea is to eliminate the highest order at each step. 
A: $$\left(x^m-\dfrac1{x^m}\right)\left(x^2+\dfrac1{x^2}\right)=x^{m+2}-\dfrac1{x^{m+2}}+x^{m-2}-\dfrac1{x^{m-2}}$$
$$\implies F_{m+2}=F_m\left(x^2+\dfrac1{x^2}\right)+F_{m-2}$$  where $F_r=x^r-\dfrac1{x^r}$
Now  $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=F_1^2+2$
$$F_3=\left(x-\dfrac1x\right)^3+3\left(x-\dfrac1x\right)$$
$$F_5=\left(\left(x-\dfrac1x\right)^2+2\right)F_3+F_1$$
