If $A$ and $B$ are linear transforms then show that My attempt:$$
 A(AB -BA) = (AB-BA )A\Leftrightarrow
2ABA = A(AB) +  (BA)A\Leftrightarrow
2ABA = AAB + BAA
$$
Now expanding the required to prove part for ($k= 2$)
$$
AAB- BAA = 2A(AB-BA )\Leftrightarrow
AAB - BAA = 2AAB - 2ABA\Leftrightarrow
2ABA - BAA = AAB 
$$
now using 1. We get $\text{LHS}=\text{RHS}$ 

Now how to extend it further ?
You can use induction. For $k=1$, there's nothing to do. Now, let $k\in\mathbb N$ and suppose that $A^kB-BA^k=kA^{k-1}(AB-BA)$. Then\begin{align}A^{k+1}B-BA^{k+1}&amp;=A^{k+1}B-A^kBA+A^kBA-BA^{k+1}\\&amp;=A^k(AB-BA)+(A^kB-BA^k)A\\&amp;=A^k(AB-BA)+kA^k(AB-BA)\\&amp;=(k+1)A^k(AB-BA).\end{align}
As you proved k=2.  Now assume for k=n. And try to prove for k=n+1 by using (AB -BA)A= A(AB -BA) . Hence by principle of mathematical induction it holds for all k.