I'm trying to work out different product rules for the Laplacian and I've gotten stuck on the Laplacian of a cross product.
Suppose $A = A_i\mathbf{\hat{e}}_i$ and $B = B_i\mathbf{\hat{e}}_i$ are vectors in $\mathbb{R}^3$. Then
\begin{align} \Delta\left(A\times B\right) &= \epsilon_{ijk}\Delta\left(A_jB_k\right)\mathbf{\hat{e}}_i \\ &= \epsilon_{ijk}\left[A_j\Delta B_k + 2\partial_mA_j\partial_mB_k + B_k\Delta A_j\right]\mathbf{\hat{e}}_i \\ &= A\times\left(\Delta B\right) + 2\epsilon_{ijk}\partial_mA_j\partial_mB_k\mathbf{\hat{e}}_i + \left(\Delta A\right)\times B \end{align}
I can't identify the second term on the last line with anything meaningful in terms of standard vector calculus operations.
We use the formula
\begin{equation}\tag{1} \Delta(fg) = f\Delta g + 2\nabla f \cdot \nabla g + g\Delta f \end{equation}
in the deriving an expression for $\vec{X}=\Delta(\vec{A} \times \vec{B})$.
Now,
\begin{equation}\tag{2} \vec{A} \times \vec{B} = \hat{e}_x(A_yB_z - A_zB_y) + \hat{e}_y(A_zB_x - A_xB_z) + \hat{e}_z(A_xB_y - A_yB_x) \end{equation}
so that, using the first equation,
\begin{eqnarray} X_x &=& (A_y\Delta B_z - A_z\Delta B_x) + 2(\nabla A_y \cdot \nabla B_z - \nabla A_z\cdot\nabla B_y) + (B_z\Delta A_y - B_y\Delta A_z) \nonumber \\ X_y &=& (A_z\Delta B_x - A_x\Delta B_z) + 2(\nabla A_z \cdot \nabla B_x - \nabla A_x\cdot\nabla B_z) + (B_x\Delta A_z - B_z\Delta A_x) \nonumber \\ X_y &=& (A_x\Delta B_y - A_y\Delta B_x) + 2(\nabla A_x \cdot \nabla B_y - \nabla A_y\cdot\nabla B_x) + (B_y\Delta A_x - B_x\Delta A_y) \nonumber \end{eqnarray} The manipulations are a little tedious but you can use "cyclic permutations" trick for getting successive terms of a cross product. The first terms on the right hand side of the three equations are just the components of $\vec{A} \times \Delta\vec{B}$. Likewise, the third terms of are the components of $\vec{B} \times \Delta\vec{A}$. Therefore,
\begin{equation}\tag{3} \vec{X} = \vec{A}\times\Delta\vec{B} + \vec{Y} + \vec{B}\times\Delta\vec{A} \end{equation}
where
\begin{equation}\tag{4} \frac{\vec{Y}}{2}=\hat{e}_x(\nabla A_y \cdot \nabla B_z - \nabla A_z\cdot\nabla B_y) + \hat{e}_y(\nabla A_z \cdot \nabla B_x - \nabla A_x\cdot\nabla B_z) + \hat{e}_z(\nabla A_x \cdot \nabla B_y - \nabla A_y\cdot\nabla B_x) \end{equation}
The term $\partial_m A_j \partial_m B_k$ is indeed a dot product of gradient of $A_j$ and the gradient of $B_k$ and $\epsilon_{ijk}$ ensures that $j \ne k$. Unless I have made a mistake in the derivation the middle term in your equation is just the vector $\vec{Y}$ defined in equation (4).