<s>[INST] $x^3 = x$ for all $x \in R$, where $R$ is a ring. Prove that $6x = 0$ for all $x \in R$. I am very confused about this problem. Because, clearly 6x = 0  only has one solution, 0. But I can make 

$x^3 = x$      
$x^3 -x=0$   (Adding additive inverse of x to both sides)
$x(x^2-1)=0$ (Distributive property of the ring)
$x(x+1)(x-1) = 0$ (Distributive property of the ring again)

which gives $x=0,1,-1$ as solutions. 
Not also I don't really know how to prove this problem, it confuses me how this is possible? Is it because the ring could not be an integral domain?
 [/INST] $(x+1)^3=x^3+3x^2+3x+1=x+1=x+3x^2+3x+1$implies that $3x^2+3x=0$. $3(x^2+x)=0$.
$3((x+1)^2+x+1)=3(x^2+2x+1+x+1)=3(x^2+x)+6x+6=0$.
$(1+1)^3=1+1$, $8=2$ implies $6=0$, so $6x+6=6x=0$.
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