Give the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$

Question

I need to give the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$ and show that $P_n(X)=(2n+1)\prod_{k=1}^{n}(X^2-\frac{1}{t_{n,k}^{2}})$ with $t_{n,k}^{2}=\tan(\frac{k\pi}{2n+1})$.

Here is what I have done so far: $$P_n(X)=\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{2k}X^{2k}=(2n+1)\left(X^{2n}+(-1)\frac{(1-2n)}{3}X^{2n-2}+...+(-1)^{n-1}nX^2+\frac{(-1)^n}{2n+1}\right)$$

but now i don't know how to progress further to get the asked result.

Answer 1

Please see here for the solution.

Also, please see the discussion here.