Solve $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$

Note that $x=-1$ is not a solution. This is a GP with common ratio $r=(x-1)/(x+1)\ne1$ and the sum is$$(x+1)^{63}\left[\frac{r^{64}-1}{r-1}\right]=0\implies r=\pm1$$Since $r\ne1$ it must equal $-1$ giving only real solution as $x=0$.

$$\frac {a^n - b^n}{a-b} = a^{n-1} + a^{n-2} b + a^{n-3}b^2 + \dots + ab^{n-2} + b^{n-1}$$

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, $\pm 1$).

You can solve it by using the formula of GP
Here $a = (x+1)^{63}$ and $r = \dfrac{x-1}{x+1}$