I want an example to show that if $a,b$ are nilpotent elements of a ring $R$ with 1 and if $c$ is any element of $R$, then $abc=0\Rightarrow acb=0$ but $cab=0$ does not imply $acb=0$.
This is unlike symmetric ring, where we know that if $a,b,c\in R$ and $abc=0$ implies that $acb=0$.
Please help me to find a ring where to search for an example or help me to show that if $abc=0 \Rightarrow acb=0$, then $cab=0 \Rightarrow acb=0$ for all nilpotent elements $a,b$ in $R$ and for all $c\in R$.
Cross-posted on MathOverflow: http://mathoverflow.net/questions/144485/example-of-a-ring-satisfying-this-variant-definition-of-symmetric-on-nilpotent
Edit: The question has received an answer at MO, which looks correct to me -- Todd Trimble.