TITLE: If binomial expansion holds for $(A+B)^n$, does it follow that $A$ ad $B$ commute?
QUESTION [6 upvotes]: Let $A, B$ be $2 \times 2$ real matrices. Is $AB=BA$ if $(A+B)^3=A^3+3A^2B+3AB^2+B^3$?
Also, generalization for arbitrary positive integer $n$ is appreciated.

This is not a homework problem, but something that interests me.

REPLY [3 votes]: I have no idea for the general case, but for cubic power, the answer is no. Here is a counterexample:
$$
A=\pmatrix{0&1\\ 0&1},\ B=\pmatrix{1&-1\\ 2&-2},\ AB-BA=\pmatrix{2&-2\\ 2&-2}\ne0.
$$
Both $(A+B)^3$ and $A^3+3A^2B+3AB^2+B^3$ are equal to $\pmatrix{1&0\\ 2&-1}$.