Prove that in a ring with $x^3 = x$ we have $x+x+x+x+x+x=0$. This was an exercise on a course on abstract algebra at the University of Groningen. I have been working on this for ages, but I can't seem to figure it out.
Problem
Let $R$ be a ring with $\forall x \in R: x^3 = x$.  Prove that $x+x+x+x+x+x=0$.
Tried
If $x=0$, the statement is of course trivial.
If $x \neq 0$, we have $x(x^2-1)=0=(x^2-1)x$, so either $x^2 = 1$, so $x$ is a unit, or $x^2-1$ and $x$ are zero-divisors.
And this is as far as I get... Any help would be appreciated! :)
Edit: guide to answer
The following hint was provided by Abel:
$$\begin{align}
x \in R &\implies x+x \in R\\
&\implies x+x=(x+x)^3
\end{align}$$
If you then work out all the terms, the answer will follow quickly.
 A: When examining the consequences of an identity in an algebra, the natural place to start is the ground, i.e. the ground terms generated by evaluating the identity at the constants of the algebra. For rings we have constants $0$ and $1$ so its natural to look first at what the identity implies in the subring generated by these constants. In our example, we have the identity $\rm\ f(x) = x^3\!-x = 0,\ $ so we deduce that $\rm\:f(0)=0,\ f(1)=0,\ \color{#C00}{f(2) = 6},\ f(3) = 24,\,\ldots$ are all zero. Thus $\rm\:\color{#C00}6 = 0\:\Rightarrow\:6x = 0.\:$  
The proof generalizes to  rings without $1$, simply evaluate $\rm\:f\:$ at $\rm\:\color{#C00}{2x} = x\!+\!x\:$ instead of at $\rm\:\color{#C00}2.\:$ In fact we can deduce further identities using polynomial arithmetic on the identities. For example, we easily infer $\rm\: 0 = f(x\!+\!1)-f(x) = 3\,(x^2\!+\!x).\:$ More generally we can compute gcds/resultants of $\rm\:f(x\!+\!n),\,f(x),\:$ and perform more complex eliminations using multivariate generalizations of the Euclidean algorithm (e.g. Grobner bases). These techniques come in handy when attacking more difficult problems, e.g. Jacobson's commutativity theorem: $\rm\: x^{n} = x\:\Rightarrow\:xy = yx$.
A: Try $x+x = (x+x)^3$ and expand the latter.
