Easy Algebra Manipulation

$a^6-b^6=(a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)$

$\frac{a^6-b^6}{a-b}=a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$

Let $a=x+1$ and $b=x-1$ :

$\begin{aligned} &&\frac{(x+1)^6-(x-1)^6}{(x+1)-(x-1)} \\ &=&(x+1)^5+(x+1)^4(x-1)+(x+1)^3(x-1)^2 \\ && + (x+1)^2(x-1)^3+(x+1)(x-1)^4+(x-1)^5=0 \end{aligned}$

$\frac{(x+1)^6-(x-1)^6}{2}=0$

$3x^5+10x^3+3=0$

$x(3x^2+1)(x^2+3)=0$

$x=\{0,\pm\frac{i}{\sqrt{3}},\pm i\sqrt{3}\}$

Our answer is the sum of all the solutions:

$0+\frac{i}{\sqrt{3}}-\frac{i}{\sqrt{3}}+i\sqrt{3}-i\sqrt{3}=\boxed{0}$

We know that $a^6-b^6=(a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)$

put $a=x+1$ and $b=x-1$

so : $(x+1)^5+(x+1)^4(x-1)+(x+1)^3(x-1)^2+(x+1)^2(x-1)^3+(x+1)(x-1)^4+(x-1)^5=0$

$\implies \frac{(x+1)^6-(x-1)^6}{x+1-x+1}=0$

$\implies (x+1)^6=(x-1)^6$

$\implies \begin{cases} & x+1=x-1 \text{ impossible } \\ & x+1=-x+1 \Rightarrow x=0 \end{cases}$