Solve the following equation for x : $$(x + 1)^{63}  +  (x + 1)^{62}(x−1)  +  (x + 1)^{61}(x−1)^{2} + . . .+(x−1)^{63}= 0$$
My approach was:
Its a GP with 
$$r= \frac{(x-1)}{(x+1)}$$
Then with the expression:
$ ar$$n-1$$=(x-1)$$63$
Plugging value of r and a it results to it:
$n=64$
Plugging in GP sum formula i finally get to this :
$(x-1)$$64$ - $(x+1)$$64$ =$0$
Then what should be done 
Solve for all 64 values? 
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 A: You have to solve for
$$\biggl(\frac{x+1}{x-1}\biggr)^{\!64}=1,\quad x\ne 1$$
So set $u=\dfrac{x+1}{x-1}$ and solve $\;u^{64}=1$:
$$u=\mathrm e^{\tfrac{ik\pi}{32}},\enspace\text{whence}\enspace x=\frac{u+1}{u-1}=-i\cot\frac{k\pi}{64}\quad (1\le k\le63).$$
The only real solution is $x=0$, and it corresponds to the case $k=32$.
A: Note first of all that $x=-1$ is not a solution. This allows you to write the equation as:
$$\left(\frac{x+1}{x-1}\right)^{64}=1.$$
There are 64, 64th roots of unity, given by $\zeta^k$ for $\zeta$ a principal root and $k=0,1,\dots,63$.
This gives 
$$x_k=\frac{1+\zeta^k}{1-\zeta^k}.$$
You might be concerned that $x_k$ could be real even if $\zeta^k$ isn't.
It is an exercise to show that this can't happen (assume $(1+z)/(1-z)\in \mathbb{R}$ and show that $z\in \mathbb{R}$).
Therefore $\zeta^k=\pm 1$. 
Clearly $\zeta_k=1$ doesn't work and so we must have $\zeta^k=-1$ and so $x=0$.
A: Note that we have $$(x+1)^{64}=(x-1)^{64}$$
From the Geometric Progression. 
However, note that as $t^{64}$ is a function that is increasing if $t>0$, but decreasing if $t<0$, we have that $$(x+1)^{64}=(x-1)^{64} \iff (x+1)= \pm (x-1)$$
So $x=0$ as $x-1 \neq x+1$. 
A: It gives $x-1=-(x+1)$ and $x=0$ or $x-1=x+1$, which is impossible.
