<s>[INST] Let $G$ be a group such that for all $x,y\in G,(xy)^3=x^3y^3$ and $(xy)^8=x^8y^8$. Prove that $G$ is Abelian or not? I tried to prove it like $(xy)^3=xyxyxy=x^3y^3 $ which implies
$$ x^{-1}/x^3y^3=xyxyxy$$
$$x^2y^3=yxyxy/y^{-1}$$
$$ x^2y^2=yxyx $$
and I get $x^2y^2=(yx)^2$.
But I dont know how to prove that $(xy)^3=x^3y^3$
 [/INST] I prove commutativity. (The comments
reveal that a more general result was proved in 2017.)  The proof is for cancellative semigroups.
$$\ (xy)^3=x^3y^3\tag{A}$$
$$\ (xy)^8=x^8y^8\tag{B}$$
Using cancellation on (A), $(yx)^2=x^2y^2 \tag 1$
Therefore, $x^2y^3x=(yx)^3=y^3x^3$ so that $x^2y^3=y^3x^2\tag 2$
Using cancellation on (B), $(yx)^7=x^7y^7 \tag 3$
Therefore, $x^7y^8x=(yx)^8=y^8x^8$ so that $x^7y^8=y^8x^7\tag 4$
Apply (2) to (4) to obtain $xy^8x^6=y^8x^7$ and cancel to get
$xy^8=y^8x \tag 5$
From $y^6x^2y^2=x^2y^8=y^8x^2,$ we obtain
$x^2y^2=y^2x^2 \tag 6$
Apply (6) to (2) to obtain the powerful identity
$x^2y=yx^2 \tag 7$
Commutativity now follows by cancellation on
$$x^3y^3=(xy)^3=x(xy)^2y=x^2yxy^2.$$
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