My problem is basically how do I find the roots of $(x+1)^n+(x-1)^n=0$
It is from an exercise list of my professor. I tried open it in a sum and all I got was
if n is odd, then $x=0$ is a root and I get something like $2x(x^{n-1}+ax^{n-3}+\cdots + 1)$ with their respective binomial coefficients.
And if n is even, then i got almost the same thing but without the $x$.
Just write it so: $$\left(\frac{x-1}{x+1}\right)^n=-1$$ and solve the equation $z^n=-1$ or $$z^n=1\left(\cos\pi+i\sin\pi\right).$$