G is a group and $(ab)^3=a^3b^3$ for all $a,b \in G$. Prove (or disprove with a counterexample) that if $(ab)^3=(ba)^3$, then $ab=ba$.

Question

Proposition. Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a,b \in G$. If $(ab)^3=(ba)^3$, then $ab=ba$.

Is it true or false? So far I've only been able to prove that powers of $a$ commute with $b^3$ and powers of $b$ with $a^3$.

Answer 1

Hint:

Let $U(3, \mathbb F_3)$ be the group of all $3 \times 3$ upper triangular matrices with all diagonal entries $1$, over the field $\mathbb F_3$. Thus, the elements are all the matrices of the form $$\begin{bmatrix}1 & a & b\\0 & 1 & c\\0 & 0 & 1\end{bmatrix}$$ where $a, b, c \in \mathbb F_3 = \{0,1,2\}$, the field of order $3$.

1. What is the exponent of the group [the least positive $n$ such that $g^n = 1$ for all group elements $g$]? Or: Determine the order of each element.
2. Is the group Abelian?
3. What do Points 1 and 2 imply about the status of the Proposition in this group?