TITLE: If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative.
QUESTION [54 upvotes]: Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring.

REPLY [4 votes]: Here goes a proof that provides motivation to find a method for the more general theorem of Jacobson:

If $R$ is a ring such that for every $a\in R$ there exists $n(a)>1$ such that $a^{n(a)}=a$, then $R$ is commutative.

This method is clearly a big overkill for the case $n(a)=n=3$, since it relies heavily on the structure of rings, but as I said it has pedagogical value:
First observe that for a division ring $D$ the conclusion is rather easy, since the commutativity of $D$ arises from the reversion of the order of the product when we apply inverses:
$$a^3=a\Rightarrow a^2=1\Rightarrow a^{-1}=a \ \ \ (a\neq 0)$$
$$ab=(ab)^{-1}=b^{-1}a^{-1}=ba  \ \ \ (a,b\neq 0)$$
Of course, trivially $0\cdot a=a\cdot 0$ for every $a\in R$.
So a natural question is: can we reduce our problem to that of division rings? In other words: can our ring be written from a family of division rings in such a way that
1) the division rings satisfy $a^3=a$, and
2) our ring inherits commutativy from them?
A reasonable answer is that 1) would be satisfied if $R$ is epimorphic on every division ring $D_i$, while 2) would be satisfied if $R$ is injected in the sum or product of the $D_i$. 
Hence we see that we if $R$ is a subdirect product of division rings then we can get our desired scheme of proof.
Now the following reasoning requires some experience: as we know, primitive rings are strongly related to division rings (via Schur's lemma, the Jacobson density theorem, etc.). Can we at least prove that our ring is a subdirect product of primitive ones?
In that case, $R$ would be semiprimitive, and we know that conversely every semiprimitive ring is a subdirect product of primitive ones. So, is $R$ semiprimitive?
We need to show that the Jacobson radical $J(R)$ is $0$. We resort to the characterization of $J(R)$ by elements. If $a\in J(R)$ then $a^2\in J(R)$ and $1-a^2$ is invertible, but we know $a^3=a\Rightarrow a(1-a^2)=0$, so $a=0$.
Therefore $R$ is a subdirect product of primitive rings $R_i$. Does the inherited identity $a^3=a$ force $R_i$ to be a division ring? 
What can be said in special cases? If $R_i$ is artinian, then $R_i=$Mat$_n(D)$ for some $n\in\mathbb{N}$ and some division ring $D$. If $n>1$ then there are nonzero nilpotent elements in $R_i$, that cannot satisfy $a^3=a$. Therefore $n=1$ and $R_i=D$.
What if $R_i$ is not artinian? Then by a consequence of the Density theorem (Litoff's theorem for primitive rings) we have there exists a division ring $D$ such that for every $n\in\mathbb{N}$ there is an epimorphism from $R_i$ to Mat$_n(D)$. So the same argument as before applies: $a^3=a$ in $R_i$ would imply $a^3=a$ in Mat$_n(D)$ for every $n$, but that is impossible, so $R_i$ is artinian and therefore a division ring.
We have arrived at the end of our proof. Now, if we want to prove Jacobson's theorem in full generality, it is natural to try the same scheme of proof (due to its degree of abstraction) and indeed it works, the only difficulty being in showing that $a^{n(a)}=a$ forces a division ring to be a field. This is how Jacobson's commutativity theorem is proved by Herstein in chapter 3 of Noncommutative rings (second edition, 1971).