If $a \ne b$ in a ring $R$ satisfy $a^3 = b^3$ and $a^2b = b^2a$, show that $a^2 + b^2$ is not a unit. If $a \ne b$ in a ring $R$ satisfy $a^3 = b^3$ and $a^2b = b^2a$, show that $a^2 + b^2$ is not a unit.
So I am thinking that I should be able to do this by contradiction. So if I assume there is some element $z\in R$ such that $(a^2 + b^2)z=1$ then 
$a^2z + b^2z=1$, but I'm not sure this helps in any way. 
I also tried this just to see if it would reveal the right way to proceed, but I am still stuck.
$a^2 + b^2=a^2 + b^2$
$(a^2 + b^2)b=(a^2 + b^2)b$
$a^2b + b^3=a^2b + b^3$
$a^2b + b^3=b^2a + a^3$

You have, $(a-b)(a^2+b^2)=a^3+ab^2-ba^2-b^3=0$
