If $ab - ba = c, bc - cb = a, ca - ac = b$ in a ring, prove that $a^2 + b^2 + c^2$ commutes with $a, b$ and $c$.

Question

I am stuck with a problem about a non-commutative ring. (I am rather new with abstract algebra.) By only putting $a, b, c$ in their expanded forms into equation
$a (a^2 + b^2 + c^2) = (a^2 + b^2 + c^2) a$
I was not getting anywhere. How can one solve a problem like this? Having three equations for three variables seemed like I could calculate them, but I was not able to.

The problem is the following:

Let $R$ be a ring. For elements $a, b, c$ form $R$ is $ab - ba = c, bc - cb = a, ca - ac = b$
Prove that $a^2 + b^2 + c^2$ commutes with $a, b$ and $c$.

It obviously holds for the trivial solution that is $a = b = c = 0$. But I can not prove this for all $a, b, c$. It looks like $a, b$ and $c$ has something in common, might be identical, but can not prove it nor use it in proving commutativity.

Answer 1

I mean why not just look a little further? $$a(a^2+b^2+c^2) - (a^2+b^2+c^2)a = ab^2+ac^2 - b^2a - c^2a=$$ $$ = (ba+c)b+ac^2 - b^2a - c^2a = bab+c(b -ca)-b^2a+ac^2= $$ $$ = b(ba+c)-cac-b^2a+ac^2 = bc-cac+ac^2 = (b-ca+ac)c=0$$

Answer 2

$$aa^2=(aaa)=a^2a$$ $$ab^2=(ab)b=(ba+c)b=bab+cb=b(ab)+cb=b(ba+c)+cb=b^2a+bc+cb=b^2a+(cb+a)+cb$$ $$ac^2=(ac)c=(ca-b)c=cac-bc=c(ac)-bc=c(ca-b)-bc=c^2a-cb-bc=c^2a-cb-(a+cb)$$

Now add the three equations.

Answer 3

We have that $[a,b]=c$, $[b,c]=a$ and $[c,a]=b$. Then: $$\begin{align} [a^2+b^2+c^2, a]&=[a^2,a]+[b^2,a]+[c^2,a]\\ &=b[b,a]+[b,a]b+c[c,a]+[c,a]c\\ &=b[b,a]+[b,a]b+cb+bc\\ &=b(-[a,b])+(-[a,b])b+cb+bc\\ &=b(-c)+(-c)b+cb+bc\\ &=0 \end{align}$$ And you can do the same for the other $2$ cases.