Problem. Let G be a finite group whose order is not divisible by 3. Suppose that (ab)3 = a3b3 for all a, b in G. Prove that G must be abelian. Attempt. I know that G has no subgroups of order 3, hence no elements of order 3. Thus, if (ab)3 = a3b3 = e, then ab = e right? I don't know how to proceed. Any tips?