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.)

You can always pick a set $X$, consider the free $\mathbb Z$-algebra $A=\mathbb Z\langle X\rangle$, and divide by the ideal generated by all the elements $x^3-x$ for $x\in A$ to que a ring $B_X$, no?
(Since the quotient is going to be commutative, you can also start from the polynomial ring...)
To show that the resulting quotient is non-trivial, consider the many maps $B_X\to\mathbb Z_3$, which you get from functions $X\to\mathbb Z_3$. Indeed, for every example of a ring $R$ satisfying the condition, you can pick an $X$ and a surjective map $B_X\to R$. So these examples are universally complicated :)
Later: Notice that $\mathbb Z_6$ is an example of a ring satisfying the identity, so $B_X$ is not a product of copies of $\mathbb Z_3$.
