Proving G is Abelian: A Group Theory Case

mansi
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seeing lots of group theory here after a really long time...
let G be a finite group of order n, where n is not divisible by 3. suppose
(ab)^3 = a^3 b^3 ,for a, b in G . prove that G is abelian.
 
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i guess i must( or rather i am supposed to) add my thought process...
since 3 does not divide n (n,3)=1 so we have un +3v=1...now for any x in G, x = x^(un+3v)= x^3v.
so G ={y^3 / y is in G}...using this i think we're supposed to show that ab=ba for all a in G. here's where I'm stuck.
 
It's frequently useful to reverse what's inside an exponent. For example:

b * (ab)^3 * a = b(ababab)a = (ba)^4
 
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The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

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