I want to prove the following identity: For $\mathbf a,\mathbf b\in \mathbb R^3, M\in M_3(\mathbb R)$, we have $$\mathbf b\times(\mathbf aM)-\mathbf a\times(\mathbf bM)=(\mathbf a\times\mathbf b)M-(\mathrm{tr} M )(\mathbf a\times\mathbf b)$$
One direct way to prove it is to write out the components of $\mathbf a,\mathbf b$ and $M$, then compare the result of $LHS$ and $RHS$ by brute force, but this way seems not so beautiful. Is there any nice (elegant) way to prove this identity? I guess probably there's a geometric interpretation of this identity.
The proposed identity may be written as $$(\epsilon_{ijk}\delta_{ml} -\epsilon_{imk}\delta_{lj} -\epsilon_{lmj}\delta_{ki} +\epsilon_{imj}\delta_{lk})a_mb_jM_{lk}=0$$ where the expression in brackets is almost one of the identities listed in equations (5)-(9) in "The Isotropic Invariants of Fifth-rank Cartesian Tensors", Int. J. Quantum Chem. vol V 381-386 (1971).
Is the author of the question certain the first term on the right hand side of their identity is not $M \cdot ({\bf a}\times{\bf b})$, rather than $({\bf a}\times{\bf b})\cdot M$? (Assuming $M$ is not symmetric.) If it were, then the third term in the brackets would be $\epsilon_{kmj}\delta_{li}$, which yields a more symmetric result, and the expression in the brackets does vanish!