Let $\displaystyle G$ be a group. Suppose that

$\displaystyle (\alpha\ast\beta)^3=\alpha^3\ast\beta^3$ -----(1)

for all $\displaystyle \alpha, \beta \in G$.

(i)Show that for all $\displaystyle \alpha, \beta \in G$, then

$\displaystyle (\beta\ast\alpha)^2=\alpha^2\ast\beta^2$ and $\displaystyle \alpha^3\ast\beta^2=\beta^2\ast\alpha^3$.

(ii)If $\displaystyle G$ satisfies (1) and the condition that the identity element $\displaystyle e$ in $\displaystyle G$, is the only element in $\displaystyle G$ satisfying $\displaystyle x^3=e$, show that for all $\displaystyle \alpha, \beta \in G$,

$\displaystyle \alpha^2\ast\beta=\beta\ast\alpha^2$

and conclude that $\displaystyle G$ is abelian.