Suppose that $(ab)^3 = a^3 b^3$ for all $a, b \in G$. Prove that G must be an abelian goup [Group Theory].

First you need to establish that every element of the group is a cube. Since the group has order not divisible by 3, we know that if $?^3=?$ then $?=?$. Now suppose that $?^3=?^3$. Then then we have $?=(?^3)(?^{?1})^3=(??^{?1})^3$ so that $??^{?1}=?$ whence $?=?$. This means that no two cubes of different elements are equal. So if we cube all $?$ elements in the group we get $?$ different results. Hence every element of the group must be a cube.

Now consider $(???^{?1})^3$ in two ways. Writing it out in full and cancelling $??^{?1}=?$ we get $??^3?^{?1}$. Using the special relation we have for the group we get $(??)^3(?^{?1})^3=?^3?^3(?^{?1})^3$. Setting these equal and cancelling: $?^3=?^2?^3(?^{?1})^2$ or $?^3?^2=?^2?^3 $.

Since $?$ and $?$ were completely arbitrary, every square commutes with every cube. But every element of the group is a cube, so every square commutes with everything. Now we can write $ ??????=(??)^3=?^3?^3 $ and when we cancel and use the fact that squares commute we find that $????=?^2?^2=?^2?^2$, hence $??=??$.