I wish to show that
\begin{equation} \nabla \cdot (A \times B) = B \cdot(\nabla \times A)- A \cdot(\nabla \times B), \end{equation}
using component notation. Here is what I have thus far
\begin{equation} \begin{split} \nabla \cdot (A \times B) =& \nabla \cdot (a_i b_j \epsilon_{ijk} \hat{e_k}) \\ =& \partial_m (a_i b_j \epsilon_{ijk} \hat{e_k})_m. \end{split} \end{equation} I am stuck after this step. I don't know how to 'introduce the product rule' into the bracketed component.
Any help would be greatly appreciated!
EDIT: There was a typo in the first equation that has been corrected.
$$ \nabla\cdot (A \times B) = \partial_k(A \times B)_k= \partial_k(a_i b_j \epsilon_{ijk})=\epsilon_{ijk}\partial_k(a_i b_j)=\\ =\epsilon_{ijk}b_j\partial_k a_i+\epsilon_{ijk}a_i\partial_k b_j=\\ =b_j(\epsilon_{kij}\partial_k a_i)-a_i(\epsilon_{kji}\partial_k b_j)=\\ =b_j(\nabla\times A)_j-a_i(\nabla\times B)_i=\\ = B \cdot(\nabla \times A)- A \cdot(\nabla \times B), $$ having used $\epsilon_{ijk}=\epsilon_{kij} $ and $\epsilon_{ijk}=-\epsilon_{kji} $.