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 $𝑥^3=𝑒$ then $𝑥=𝑒$. Now suppose that $𝑎^3=𝑏^3$. Then then we have $𝑒=(𝑎^3)(𝑏^{−1})^3=(𝑎𝑏^{−1})^3$ so that $𝑎𝑏^{−1}=𝑒$ whence $𝑎=𝑏$. This means that no two cubes of different elements are equal. So if we cube all $𝑛$ elements in the group we get $𝑛$ different results. Hence every element of the group must be a cube.

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

Since $𝑎$ and $𝑏$ 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 $ 𝑎𝑏𝑎𝑏𝑎𝑏=(𝑎𝑏)^3=𝑎^3𝑏^3 $ and when we cancel and use the fact that squares commute we find that $𝑏𝑎𝑏𝑎=𝑎^2𝑏^2=𝑏^2𝑎^2$, hence $𝑎𝑏=𝑏𝑎$.