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 above mentioned $a,b,c$.



Example of a ring satisfying this variant definition of "symmetric" on nilpotent elements
If I understand correctly, there are two conditions on the ring $R$:


(1) if $a$ and $b$ are nilpotent elements of $R$, and if $c$ is arbitrary, then $abc=0$ implies $acb=0$; 


(2) if $a$ and $b$ are nilpotent elements of $R$, and if $c$ is arbitrary, then $cab=0$ implies $acb=0$.


There exists a ring for which condition (1) is satisfied, while condition (2) fails. Indeed, let $k$ be a field, and let $R$ be the path algebra of the quiver
$$ 1 \stackrel{x}{\leftarrow} 2 \circlearrowleft^y $$
over $k$, modulo the relation $y^2=0$. Let $e\_1$ and $e\_2$ be the paths of length $0$ at vertices $1$ and $2$, respectively. I compose arrows from left to right: $xy$ is a non-zero path, while $yx$ is not.


Then any nilpotent element is a linear combination of $x$, $y$ and $xy$. Let $(ax + by+ cxy)$ and $(dx+ey+fxy)$ be two such elements, and let $(ge\_1 + he\_2 + ix + jy + lxy)$ be an arbitrary element. We have
$$(ax + by+ cxy)(dx+ey+fxy)(ge\_1 + he\_2 + ix + jy + lxy) = (aeh)xy \\
 (ax + by+ cxy)(ge\_1 + he\_2 + ix + jy + lxy)(dx+ey+fxy) = (aeh)xy.$$
Thus condition (1) is satisfied. However, we have that $e\_2xy = 0$, while $xe\_2y = xy \neq 0$. Thus condition (2) is not satisfied.


