Prove that this is an abelian group 
$G$ is a group. $f:\:G\to G,\:a\mapsto a^3$ is a group monomorphism. Prove that $G$ is abelian.

My idea is that it's sufficient to prove that $(aba^{-1}b^{-1})^3=e$. So
$$
(aba^{-1}b^{-1})^3=a^3b^3a^{-3}b^{-3}
$$
and I am stuck. Or
$$
(aba^{-1}b^{-1})^3=a^3(ba^{-1}b^{-1})^3=a^3ba^{-3}b^{-1}
$$
but this kind of effort does not seem to be of any help.

Here's my version (revised) . . .

Since the map $\:x\mapsto x^3$ is a monomorphism,
\begin{align*}
&\hspace{1pt}(ab)^3 = a^3b^3,\text{ for all }a,b \in G.\tag{1}\\[6pt]
&\hspace{1pt}\text{If }a,b \in G,\text{ and }a^3 = b^3,\text{ then }a = b.\tag{2}\\
\end{align*}
Let $a,b \in G.$
\begin{align*}
\text{Then}&&a^3b^3a^{-3}&=(aba^{-1})^3&&[\text{by} (1)]\\
&&&= (aba^{-1})(aba^{-1})(aba^{-1})&&\\
&&&= ab^3a^{-1}&&\\[8pt]
\text{But then}&&a^3b^3a^{-3} &= ab^3a^{-1}&&\\
\implies&& a^2b^3&=b^3a^2&&\\[8pt]
\text{So we have}&&a^2b^3&=b^3a^2,\,\text{ for all }a,b \in G.&&\tag{3}\\
&&&\text{(thus, squares commute}&&\\
&&&\text{ with cubes)}&&\\[8pt]
\text{Then}&&(a^2b)^3 &= a^6b^3&&[\text{by} (1)]\\
&&&= b^3a^6&&[\text{by} (3)]\\
&&&= (ba^2)^3&&[\text{by} (1)]\\[8pt]
\text{But then}&&(a^2b)^3 &= (ba^2)^3&&\\
\implies&& a^2b&=ba^2&&[\text{by} (2)]\\[8pt]
\text{So we have}&&a^2b&=ba^2,\,\text{ for all }a,b \in G.&&\tag{4}\\
&&&\text{(thus, squares commute}&&\\ 
&&&\text{ with everything)}&&\\[8pt]
\text{Then}&&(ab)^6 &=((ab)^2)^3&&\\
&&&= ((ab)(ab))^3&&\\
&&&= (ab)^3(ab)^3&&[\text{by} (1)]\\
&&&= a^3b^3a^3b^3&&[\text{by} (1)]\\
&&&=a(a^2)b^3a^3(b^2)b&&\\
&&&=ab^5a^5b&&[\text{by} (4)]\\[8pt]
\text{But then}&&(ab)^6 &= ab^5a^5b&&\\
\implies&& a(ba)^5b &=ab^5a^5b&&\\
\implies&& (ba)^5 &=b^5a^5&&\\
\implies&& (ba)^3(ba)^2 &=b^5a^5&&\\
\implies&& b^3a^3(ba)^2 &=b^5a^5&&[\text{by} (1)]\\
\implies&& a^3(ba)^2 &=b^2a^5&&\\
\implies&& a^3(ba)^2 &=a^5b^2&&[\text{by} (4)]\\
\implies&& (ba)^2 &=a^2b^2&&\\
\implies&& (ba)^2 &=b^2a^2&&[\text{by} (4)]\\
\implies&& baba &=b^2a^2&&\\
\implies&& ab &=ba&&\\[8pt]
\end{align*}
