Let $(G,*)$ be a group with $(a*b)^3=a^3*b^3$ and $(a*b)^5=a^5*b^5\forall a,b∈G$. Show that $(G,*)$ is abelian. 
Let $(G, *)$ be a group such that $(a * b)^3 = a^3 * b^3$ and $(a * b)^5 = a^5 * b^5$ for all $a, b ∈ G$. Show that $(G, *)$ is abelian.

My attempt:
I need to show $ab = ba$.
$$(ab)^3 = a^3b^3 \implies a(ba)^2b = a^3b^3 \implies (ba)^2 = a^2b^2$$
$$(ab)^5 = a^5b^5 \implies a(ba)^4b = a^5b^5 \implies (ba)^4 = a^4b^4$$
$$(ba)^4 = ((ba)^2)^2=(a^2b^2)^2 = a^4b^4$$
$$a^2b^2a^2b^2=a^4b^4$$
$$b^2a^2=a^2b^2$$
But what next?

You also have $(ba)^3=b^3a^3$, as the statement holds for all group elements. Hence, $(ab)^2=b^2a^2$, by left and right cancellation. So, by combining with your last equality, you can get $ab=ba$.
