Prove that the group $G$ is abelian if $a^2 b^2 = b^2 a^2$ and $a^3 b^3 = b^3 a^3$ 
In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$. Prove that the group $G$ is abelian. 

My approach was the following:
Let $a,b\in G$
Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds.
Now, $$\begin{align}
a^3b^3=b^3a^3 \\
\implies aaabbb=&bbbaaa\\
\implies a\cdot a^2\cdot b^2&\cdot b=b\cdot b^2\cdot a^2\cdot a\\
\implies a\cdot b^2\cdot a^2&\cdot b =b\cdot a^2\cdot b^2\cdot a\\
\implies ab \cdot ba \cdot a&b=ba\cdot ab\cdot ba
\end{align}$$
I am unable to proceed further.
What I need is a simple proof using simple theorems on groups, better if could be done using elementary properties.
 A: Hint: We can show that 
$$a^{6}b=ba^{6}$$
for all $a, b\in G$. With assumptions, we have 
$$a^{2}=b^{-2}a^{2}b^{2}\ \ \text{and}\ \ a^{3}=b^{-3}a^{3}b^{3},$$ 
and we get 
$$a^{6}=b^{-2}a^{6}b^{2}\ \ \text{and}\ \ a^{6}=b^{-3}a^{6}b^{3},$$ 
thus 
$$a^{6}=b^{-2}a^{6}b^{2}=b^{-3}a^{6}b^{3}$$ 
and the above relation implies that  $a^{6}b=ba^{6}$ for all $a, b\in G$. 
Proof. The center of a group $G$, denoted $Z(G)$ is the set of elements that commute with every element of $G$, which is subgroup and $Z(G)=G$ iff $G$ Abelian group.. With the above hint we get for any $a\in G$, $a^{6}\in Z(G)$, now  $a^{8}b^{6}=b^{6}a^{8}$ and so 
$$aa^{7}b^{6}=b^{6}a^{7}a;$$
therefore $a^{7}b^{6}\in Z(G)$ and since $Z(G)$ is a subgroup thus $a^{7}, a^{6}\in Z(G)$ (note that $a^{6}\in Z(G)$) and so $a^{7-6}=a\in Z(G)$ and this implies that $Z(G)=G$ i.e., $G$ is Abelian group.
A: Hint: If you can somehow show that $a^2b^3=b^3a^2$. First take $(a^2b^3)^2$ and then prove it equal to $(b^3a^2)^2$. After this show that $a^2b^3=b^3a^2$ implies the group to be abelian
