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.

Answer

To begin with

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

Therefore 6x=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
and

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

Adding we get

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

Multiply the last relation by x on the left and right to get

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 \ .

We then show that 3( 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) \ .

In particular

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

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

Attribution
Source : Link , Question Author : Aj I , Answer Author : WDR

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