If $G$ is a group whereby $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show $G$ is abelian. If $G$ is a group in which $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian.
Proof: Let $x$ be the smallest of the 3 consecutive integers. Then, we have 
$(1)(a\cdot b)^{x} =a^x\cdot b^x$,
$(2)(a\cdot b)^{x+1} =a^{x+1}\cdot b^{x+1}$ and
$(3)(a\cdot b)^{x+2} =a^{x+2}\cdot b^{x+2}$.
Using $(2)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $baba...ba = a^{x}b^{x}(4)$ whereby there are $x$ number of $a$ and $x$ number of $b$ on both sides.
Using $(1)$ and multiplying both sides on the right by $ab$, we have $\overbrace{abab...ab}^{(x+1)ab} = (a^{x}b^{x})ab (5)$.
Substitute $(4)$ into $(5)$, we get $\overbrace{ab...ab}^{(x+1)ab} =\overbrace{ba...ba}^{(x)ba}ab(6)$.
Using $(3)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $\overbrace{ba...ba}^{(x+1)ba} = a^{x+1}b^{x+1} = \overbrace{abab...ab}^{(x+1)ab}(7)$.
Combining $(6)$ and $(7)$, and multiply $a^{-1}b^{-1}...a^{-1}b^{-1}$ on the left, we get $ab = ba$. Hence $G$ is abelian.

Use $\backslash$overbrace{below}^{above}, as in (right click and select to see LaTeX commands):
$$
\overbrace{a\ldots a}^{27}
$$
The proof looks fine to me.
