My proof of this problem looks clunky...suggestions for cleanup or different approaches are welcome. Sorry, I really need to get laTeX, this looks ridiculous without it.
From 2nd ed. of I.N. Herstein's Topics in Algebra:
Let G be a finite group whose order is not divisible by 3. Suppose that (ab)^3=a^3b^3 for all a,b in G. Prove that G must be abelian.
Denote the order of G by o(G) and G's identity element by e.
We first show that there can be no a in G other than e for which a^3=e. Since if G is a finite group and a is in G, then a^o(G)=e (Corollary 2, page 43, a consequence of the definition of the order of a group element). Then a^o(G)=e=a^3n for any integer n>0. But if 3 does not divide o(G) then o(G)=3m+r, where 3 does not divide m>0 and 0<r<3, m,r integers. Thus a^o(G)=a^(3m+r)=a^(3m)a^r=a^r, one of the two distinct powers of a which do not equal a^0=a^3=e, contradicting a^o(G)=e.
Now for all a,b in G, (ab)^3=ababab=a^3b^3. Cancellation shows baba=a^2b^2. The equation must be true when a and b are not each other's inverse and when a does not equal b does not equal e (that is, a, b and e are distinct from each other). But any association on baba (namely, (b)(aba), (ba)(ba), and (bab)(a)) forces a=b. The equation can only be true for all a,b in G if all a,b in G can be commuted. Thus G is abelian.