If $a^3 b = ba^3$ and if $ a $ has order 7, show that $ab = ba$

Another approach that doesn't involve going through the inverse: since $a^7=e$ we can write $ab = a(a^7)(a^7)b = a^{15}b$; now $$\begin{align}ab &=a^{15}b \\ &=\left(a^3\right)^5b &\text{by factoring }a^{15}\text{ into blocks of }a^3\\ &= \left(a^3\right)^4a^3b &\text{pulling out one factor of }a^3\\ &= \left(a^3\right)^4ba^3 &\text{commuting it past }b\text{ using }a^3b=ba^3\\ &= \left(a^3\right)^3b\left(a^3\right)^2 &\text{doing the same with another factor of }a^3\\ &=\ldots \\ &= b\left(a^3\right)^5 &\text{once we've pulled all the }a^3\text{s to the right}\\ &=ba^{15} &\text{packing them together again} \\ &=ba(a^7)(a^7)\\ &= ba &\text{reversing the process that got us from }a\text{ to }a^{15}\\ \end{align}$$

The same argument proves more generally that in any group, if $a$ is of order $n$, $a^p$ commutes with $b$ and $\gcd(n,p)=1$ then $a$ itself commutes with $b$.

$ba^3b^{-1}=a^3$ so $ba^6b^{-1}=ba^3b^{-1}ba^3b^{-1}=a^3a^3=a^6$ Thus $ba^{-1}b^{-1}=a^{-1}$; so $b$ commutes with $a^{-1}$ and so with any power of $a^{-1}$ including $a=a^{-6}$

Given the equation $a^3b=ba^3$, multiply by $a^3$ on both sides to get $a^{-1}b=a^6b=(a^3b)a^3=ba^6=ba^{-1}$. Hence, $b$ and $a$ must commute.