Prove that $R$ is a commutative ring if $x^3=x$ Let $R$ be a ring satisfying: $\forall x\in R, \; x^3=x$. Prove that $R$ is a commutative ring.
 A: Some general facts (note that $R$ is not necessarily unital):
We call a ring $R$, J-ring (Jacobson ring), if for any $x \in R$ there is a natural number $n(x) >1$ s.t. $x^{n(x)}=x$. (In fact, Jacobson has proven that any J-ring is commutative, for the proof you may take a look at Non-commutative Rings written by Herestein)
Lemma 1: If $R$ be a J-ring, then $N(R)= \{0 \}$ where $N(R)$ denotes the nilradical of $R$.
Proof: Let $0\not= x\in N(R)$. Then there is a smallest natural number greater $1$ s.t. $x^m=0$. Since $R$ is a J-ring, there is an $n>1$ s.t. $x^n=x$. Let $m=nq+r$ where $0 \leq r <n$. Therefore, 
$$x^m=x^{nq+r}=(x^n)^qx^r=x^qx^r=x^{q+r}=0$$
However, $q+r<m$, which is a contradiction, since $m$ was chosen to be the smallest number satisfying $x^m=0$.
Lemma 2: Suppose that in a ring $R$, $N(R)= \{0 \}$, then any idempotent element $a$ i.e. $a^2=a$, lies in the center $Z(R)$.
Proof: Suppose that $x \in R$. Then
$$(axa-ax)^2=(axa-ax)(axa-ax)=axaaxa-axaxa-axaax+axax=axaxa-axaxa-axax+axax=0.$$
Since $N(R)= \{0 \}$, then we have $axa-ax=0 \rightarrow axa=ax$. With the same approach and by considering $(axa-xa)^2$ we will obtain $axa=xa$. Hence, $ax=xa$ and since $x$ was an arbitrary element of $R$ then $a \in Z(R)$.
Lemma 3: In a J-ring $R$, we have $x^{n(x)-1} \in Z(R).$
Proof: $(x^{n(x)-1})^2=x^{2n(x)-2}=x^{n(x)}x^{n(x)-2}=xx^{n(x)-2}=x^{n(x)-1}$. Thus $x^{n(x)-1}$ is an idempotent element of $R$ and by Lemma 1 & 2. we get the result.
In particular, in your question, $n=n(x)=3$ and $x^2 \in Z(R),$ for any $x \in R.$ Moreover
$$xy=(xy)^3=xyxyxy=x(yx)^2y=(yx)^2xy=yxyx^2y=yx^3y^2=yxy^2=y^3x=yx.$$
Exercise: The same question with $x^4=x$ for any $x \in R.$
