sum (difference) of polynomials to the power n Is there a way to simplify the expression:
$D = (f_1(\omega)+f_2(\omega))^n-(f_1(\omega)-f_2(\omega))^n$
where $n$ is a positive integer.
In this particular problem:
$f_1(\omega)=-\omega^2+2$
$f_2(\omega)=\omega \sqrt{\omega^2-4}$
Expanding $D$ for some values of $n$:
$n=1$: $\sqrt{{\omega}^{2}-4}(2 \omega)$
$n=2$: $\sqrt{{\omega}^{2}-4}(-4\omega^3 + 8\omega)$
$n=3$: $\sqrt{{\omega}^{2}-4}\left( 8\,\omega^{5}-32\,\omega^{3}+24\,\omega\right) $
$n=4$: $\sqrt{{\omega}^{2}-4}\left(-16\omega^7+96\omega^5-160\omega^3+64\omega\right)$
$n=5$: $\sqrt{{\omega}^{2}-4}\left(32\omega^9-256\omega^7+672\omega^5-640\omega^3+160\omega\right)$

By the binomial theorem 
$$
(f_1+f_2)^n -(f_1-f_2)^n = \sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}f_2^k -\sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}(-f_2)^k.
$$
Notice that since we have $(-f_2)^k$ the second sum is alternating, so every other term cancels, leaving 
$$
2\sum_{j=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2j+1}f_1^{n-(2j+1)}f_2^{2j+1}=2\left( \binom{n}{1}f_1^{n-1}f_2 + \binom{n}{3}f_1^{n-3}f_2^3 + \dots   \right)
$$
where $\lfloor\ \rfloor$ is the floor function.
