Solve $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$ I want to find the solutions of  $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$.
It is not hard to see $x=0$ is a root of the equation. but I don't know how to solve this equation in general. I can see terms of the equation looks very similar to binomial expansion of $[(x+1)+(x-1)]^{63}$ except the coefficient of each term is $1$ rather than $63\choose k $ (for $k=0,1,\cdots,63$ ). is it possible to use binomial theorem to solve the equation? (or other approaches)

$$\frac {a^n - b^n}{a-b} = a^{n-1} + a^{n-2} b + a^{n-3}b^2 + \dots + ab^{n-2} + b^{n-1}$$
Hence:
$$(x+1)^{63} + (x+1)^{62}(x-1) + \dots+ (x-1)^{63} = \frac {(x+1)^{64}-(x-1)^{64}}{(x+1)-(x-1)} = \frac12((x+1)^{64} - (x-1)^{64})$$
If the above expression is zero, we must have $(x+1)^{64} = (x-1)^{64}$. Hence $\dfrac {x+1}{x-1}$ must be one of the 64-th roots of unity (or if you are only considering real roots, $\pm 1$).
