# If a^3 =aa^3 =a for all aa in a ring RR, then RR is commutative.

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

To begin with

$$2x=(2x)^3 =8 x^3=8x \ . 2x=(2x)^3 =8 x^3=8x \ .$$

Therefore $$6x=0 \ \ \forall x6x=0 \ \ \forall x$$.

Also

$$(x+y)=(x+y)^3=x^3+x^2 y + xyx +y x^2 + x y^2 +yxy+ y^2 x + y^3 (x+y)=(x+y)^3=x^3+x^2 y + xyx +y x^2 + x y^2 +yxy+ y^2 x + y^3$$
and

$$(x-y)=(x-y)^3=x^3-x^2 y – xyx -y x^2 + x y^2 +yxy+ y^2 x -y^3 (x-y)=(x-y)^3=x^3-x^2 y - xyx -y x^2 + x y^2 +yxy+ y^2 x -y^3$$

$$2(x^2 y +xyx+yx^2)=0 2(x^2 y +xyx+yx^2)=0$$

Multiply the last relation by $$xx$$ on the left and right to get

$$2(xy+x^2yx+xyx^2)=0 \qquad 2(x^2yx+xyx^2+yx)=0 \ . 2(xy+x^2yx+xyx^2)=0 \qquad 2(x^2yx+xyx^2+yx)=0 \ .$$

Subtracting the last two relations we have

$$2(xy-yx)=0 \ . 2(xy-yx)=0 \ .$$

We then show that $$3( x+x^2)=0 \ \ \forall x3( x+x^2)=0 \ \ \forall x$$. You get this from

$$x+x^2=(x+x^2)^3=x^3+3 x^4+3 x^5+x^6=4(x+x^2) \ . x+x^2=(x+x^2)^3=x^3+3 x^4+3 x^5+x^6=4(x+x^2) \ .$$

In particular

$$3 (x+y +(x+y)^2) =3( x+x^2+ y+ y^2+ xy+yx)=0 \, 3 (x+y +(x+y)^2) =3( x+x^2+ y+ y^2+ xy+yx)=0 \,$$

we end-up with $$3(xy+yx)=03(xy+yx)=0$$. But since $$6xy=06xy=0$$, we have $$3(xy-yx)=03(xy-yx)=0$$.
Then subtract $$2(xy-yx)=02(xy-yx)=0$$ to get $$xy-yx=0xy-yx=0$$.