[Note: when I pressed "preview post," I got the error "LaTeX ERROR: Unknown error" all over, and most of the text didn't appear. However, it seems to be a problem with my own computer because the code is correct. Please let me know if I did something wrong, and how to fix this display problem. Thanks]

This is a seemingly-innocuous problem from my abstract algebra book that turns out to be getting the better of me. Let $\displaystyle R$ be a ring such that $\displaystyle x = x^3$ for all $\displaystyle x \in R$. Then $\displaystyle R$ is a commutative ring.

I've already proven that in a ring where $\displaystyle x = x^2$, we have commutativity. This was done by first proving that $\displaystyle x + x = 0$, and then I had:

$\displaystyle a + b = (a + b)^2 = aa + ab + ba + bb = a + ab + ba + b$

whence

$\displaystyle ab + ba = 0$

But $\displaystyle 0 = ab + ab$ by the lemma discussed above, so we had $\displaystyle ab + ba = ab + ab$, and so $\displaystyle ba = ab$.

I figured I would do something similar for the present problem. In fact, I have already shown that $\displaystyle a + a + a = 0$. Proceeding as before, I then evaluated $\displaystyle (a + b)^3$, but this didn't yield anything too helpful. What did work out nicely was to evaluate $\displaystyle (a + b)^3 + (a - b)^3$ which led quickly to a nice result:

$\displaystyle abb + bab + bba = 0$

In fact, it seems like I'm almost done, because if I can only show that

$\displaystyle abb = bab = bba$

then I can prove commutativity from there. It sounds right, and it is right, but regrettably I don't see the proof of the above fact. In spite of the progress I seem to have made, I'm stuck. Any ideas?