Let $ba=a^4b^3$, show that $\mathrm{ord}(a^4b)=\mathrm{ord}(a^2b^3)$

Question

Let $ba=a^4b^3$ show that $\mathrm{ord}(a^4b)=\mathrm{ord}(a^2b^3)$ where $a,b\in G$ and $G$ is a group.

I have toiled on this problem for the past 2.5 hours. I tried substitution of the elements with their expressions. I tried taking advantage of conjugation. I also tried showing commutativity of $a$ and $b$. I don't know what I am not seeing. I hope the trick is satisfying I guess since I can't figure it out.

I would appreciate any hints. I prefer hints to complete solutions. Thank you for your time.

Answer 1

Hint: You should first be able to show that, for any $x,y\in G$, $\operatorname{ord}(xy)=\operatorname{ord}(yx)$. Now, see if this is helpful along with $$bab^{-2}=a^4b,\ a^{-2}ba=a^2b^3.$$