If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative. Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring.

I happen to have come across a recent set of exercises on many of the small-$n$ cases of Jacobson's Theorem. It also happens that my solution is different than those contained in the link of @lhf above.
So we have that $a^3 = a \quad\forall a \in R$, and so $2a = (a+a)^3 =8a$, thus $6a = 0$.
Now consider the ideals $2R$ and $3R$. The intersection of $2R$ and $3R$ is trivial, as if $a \in 2R \cap 3R$, then $a = 2r = 3s$ for some $r,s$. Thus $3a = 6r = 0 = 6s = 2a$, and so $(3-2)a = a= 0$. So $2R \oplus 3R = R$. Further, if $a \in 2R$, $b \in 3R$, then $ab, ba \in 2R \cap 3R$ and so $ab = ba = 0$. So we only worry about commutativity in each ideal separately.
In $3R$, we have both $a^3 = a$ and $2a = 2 \cdot 3r = 0$ (some $r$). Then $1 + a = (1 + a)^3 = 1 + 3a + 3a^2 + a^3 = 1 + a + a^2 +a = 1 + a^2 \implies a^2 = a$. So what? In that case, we also have $(1 + a) = (1 + a)^2 = 1 + 2a + a^2 = 1 + 2a + a$, and so $2a = 0$ (yes, we have this in our ideal, but this is true in general in Boolean rings). Continuing, $(a + b) = (a + b)^2 = a^2 + ab + ba + b^2 = a + ab + ba + b$, so $ab = -ba = -ba + 2ba = ba$.
For $2R$, we have both $a^3 = a$ and $3a = 0$. Then we have that $a + b = (a + b)^3 = a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3 $$= a + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b$ on the one hand, and $a - b = (a - b)^3 = a^3 - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b^3$ $= a - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b$.
Taking the difference between these, we see $2(a^2b + aba + ba^2) = 0$, and so $a^2 b + aba + ba^2 = 0$. Multiply by $a$, and we get $a^3b + a^2ba + aba^2 = ab + (a^2b + aba)a = ab + (-ba^2)a = ab - ba = 0$. Thus $ab = ba$.
As both ideals commute separately and in products, $R$ commutes in general.
