\[ \left( {x + 1 + x - 1} \right)\left( {x + 1 - \left( {x - 1} \right)} \right)\left( {\left( {x + 1} \right)^2 - \left( {x + 1} \right)\left( {x - 1} \right) + \left( {x - 1} \right)^2 } \right)\left( {\left( {x + 1} \right)^2 + \left( {x + 1} \right)\left( {x - 1} \right) + \left( {x - 1} \right)^2 } \right) = 0 \] \[ 2x\left( {x^2 + 3} \right)\left( {3x^2 + 1} \right) = 0 \Rightarrow x = 0 \]

multiply the left by (x+1)-(x-1) and you will get (x+1)^6-(x-1)^6=0 then the answer should be x=0....

let $a=x+1$ , $b=x-1$ we get $a^5+a^4.b+a^3.b^2+a^2b^3+a.b^4+b^5=\frac{a^6-b^6}{a-b}=0$ . It shoub be easy from here .

It reminds us of the formulae $\frac{m^6–r^6}{m–r}=m^5 + m^4 r+m^3 r^2 +m^2r^3+mr^4+r^5$

u can factories 3 terms and the 3 terms , u can ezly find x = 0