Factorization of $a^{n}-b^{n}$.

Question

Let $X$ be a Banach space and let $A, B$ be two bounded linear operators on $X$ such that $AB \neq BA$. Can we write $A^{n}-B^{n}$ as follows and what is the expression of $K$: $$A^{n}-B^{n}=C(A-B)+K(AB-BA).$$ Where $C=A^{n-1}+BA^{n-2}+\dots+B^{n-1}$ and $K \in \mathbb{R}\backslash\{0\}$.

Answer 1

For $n=3$, $$A^3-B^3-(A^2+BA+B^2)(A-B)=A^2B+BAB-BA^2-B^2A$$

And you can't reasonably expect this to be a scalar multiple of $AB-BA$: there must be counterexamples with $\dim X=3$. I would guess you can make do with $A=\begin{pmatrix}1&0&0\\ 0&2&0\\0&0&3\end{pmatrix}$, $B=\begin{pmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{pmatrix}$.