Example of a ring with $x^3=x$ for all $x$ A ring $R$ is a Boolean ring if $x^2=x$ for all $x\in R$. By Stone representation theorem
a Boolean ring is isomorphic to a subring of the ring of power set of some set. 
My question is what is an example of a ring $R$ with $x^3=x$ for all $x\in R$ that is 
not a Boolean ring? (Obviously every Boolean ring satisfies this condition.)

Actually, not only would $\mathbb{Z}_3$ work, but it's the only solution that's an integral domain not of characteristic 2 (since, in such a case, $x^3-x=0\,\Rightarrow\,x\in {0,1,-1}$).  
Another solution would be $R:=\mathbb{Z}_3[\mathbb{Z}_2]$, the group ring of $\mathbb{Z}_3$ over $\mathbb{Z}_2$ (ie the ring of "polynomials" over $\mathbb{Z}_3$, except that exponents are in $\mathbb{Z}_2$).  To see why, given an element $f(x)\in R$ with
$f(x)=a_0+a_1 x^{b_1} + \cdots + a_n x^{b_n}$
with $a_i\in \mathbb{Z}_3$ and $b_j \in \mathbb{Z}_2$.  Since $R$ is a ring of characteristic 3, the Freshman's Dream implies that
$(f(x))^3=a_0^3+a_1^3 x^{3b_1} + \cdots + a_n^3 x^{3b_n} = a_0 + a_1 x^{b_1} + \cdots + a_n x^{b_n}=f(x)$.
In fact, by the same argument, if you're given any ring $T$ of characteristic 3 with $x^3=x$ for all $x\in T$, then $T[\mathbb{Z}_2]$ satisfies this property as well.
I can't think of another class of examples off the top of my head, but I'd be surprised if boolean rings and the class of examples above were the only examples of rings of this type.
EDIT:  Yes, the solution of modding out a free algebra by an appropriate ideal would also work nicely.
