Problem from Herstein on group theory The problem is:
If $G$ is a finite group with order not divisible by 3, and $(ab)^3=a^3b^3$ for all $a,b\in G$, then show that $G$ is abelian.
I have been trying this for a long time but not been able to make any progress. The only thing that I can think of is:
$$ab\cdot ab\cdot ab=aaa\cdot bbb\implies(ba)^2=a^2b^2=aabb=(\text{TPT})abba.$$
Now, how can I prove the last equality? If I write $aabb=abb^{-1}abb$, then in order for the hypothesis to be correct, $b^{-1}abb=ba\implies ab^2=b^2a$. Where am I going wrong? What should I do?
 A: Hints (remember: $\,|G|<\infty\,\,\,and\,\,\,3\,\nmid\, |G|\,$): $\,\,\forall\,\,a,b\in G\,$
$$\begin{align*}(1)&\;\;\;\text{Show that}\,\,\,\,(ba)^2=a^2b^2\\{}\\(2)&\;\;\;\text{Prove that}\;\;f:G\to G\,\,\,,\,\,f(x):=x^3\,\,,\,\,\text{is an isomorphism}\\{}\\(3)&\;\;\;\text{Define}\,\,z:=\left(aba^{-1}\right)^3 \longrightarrow \begin{cases}z=ab^3a^{-1},\;\;\;\text{and also}\\{}\\z=f(a)f(b)f(a^{-1})=a^3b^3a^{-3}\end{cases}\\{}\\(4)&\;\;\;\text{Using(2)-(3) , show that}\;\;a^2\in Z(G)\Longleftrightarrow a^2g=ga^2\,\,,\,\forall\,g\in G\\{}\\(5)&\;\;\;\text{Finally, use (1) to show that}\,\,\,ab=ba\end{align*}$$
A: *

*Suppose that $x\in G$ satisfies $ x^3=e$. Then we cannot have $x\ne e$, otherwise $x$ would have order 3, which implies that 3 divides $|G|$ (recall that the order of a group element divides the order of the group). Hence,
$$\forall x\in G:\quad x^3=e~\Longrightarrow~x=e,$$
and as $(ab)^3=a^3b^3$ for all $ a,b \in G $, we see that the function $\phi:G\to G$ defined by
$$\forall x\in G:\quad\phi(x) \stackrel{\text{def}}{=} x^3$$
is an injective group homomorphism.

*Now,
$$\forall a,b\in G:\quad ababab=(ab)^3=a^3b^3=aaabbb.$$
Hence,
$$\forall a,b\in G:\quad baba=aabb,\quad\text{or equivalently},\quad (ba)^2=a^2b^2.$$
Using this fact, we obtain
\begin{align}
\forall a,b\in G:\quad(ab)^4&= [(ab)^2]^2\\
                                  &=[b^2a^2]^2\\
                                  &=(a^2)^2(b^2)^2\\
                                  &=a^4b^4\\
                                  &=aaaabbbb.
\end{align}

*On the other hand,
\begin{align}
\forall a,b\in G:\quad(ab)^4&= abababab\\
                                  &=a (ba)^3b\\
                                  &=ab^3a^3b\\
                                  &=abbbaaab.
\end{align}

*Hence, for all $a,b\in G$, we have $aaaabbbb=abbbaaab$, which yields
$$\phi(ab)=a^3b^3=b^3a^3=\phi(ba).$$
As $\phi$ is injective, we conclude that $ab=ba$ for all $a,b\in G$.
Conclusion: $G$ is an abelian group.
A: The way, I am writing here, is from my old notes and personally I prefer the other approaches. But, maybe the given additional points below, inspire you for other problems like this problem.
We can prove that if for an integer $n$ and every $a,b\in G$, $(ab)^n=a^nb^n$, then $$(aba^{-1}b^{-1})^{n(n-1)}=e$$ The proof is easy. In fact, $$(aba^{-1}b^{-1})^{n^2}=[(aba^{-1}b^{-1})^n]^n=[a^n(ba^{-1}b^{-1})^n]^n=\cdots^*=a^nb^na^{-n}b^{-n}\\\ (aba^{-1}b^{-1})^{n}=(ab)^n(a^{-1}b^{-1})^n=a^nb^na^{-n}b^{-n}$$
*refer here for the missing steps
In your problem, we assume $G$ is not abelian, so there exist $a,b\in G, aba^{-1}b^{-1}\neq e$. According to the above lemma
$$(aba^{-1}b^{-1})^6=e$$
since we know $(ab)^3=a^3b^3$. So $|aba^{-1}b^{-1}|\big| 6$ and because of $3\nmid|G|$ so $|aba^{-1}b^{-1}|=2$. This means that $(aba^{-1}b^{-1})^2=e$. On the other hand, 
$$(ab)^3=a^3b^3\Longrightarrow (ba)^2=a^2b^2$$
(see @Haskell's answer) then $(a^{-1}b^{-1})^2(ab)^2=e$ or $(ab)^2=(ba)^2=a^2b^2$ or $ab=ba$. A nice contradiction!
