Let $$F, G:\mathbb{R}^3 \to \mathbb{R}^3$$ Prove that: $$\nabla\cdot(F\times G)=G\cdot (\nabla \times F)-F \cdot(\nabla\times G)$$
I found that:
$$∇\cdot(F×G)=(∂(F_2 G_3-F_3 G_2))/∂x+∂(F_3 G_1-F_1 G_3 )/∂y+∂(F_1 G_2-F_2 G_1 )/dz$$
and $$G\cdot(∇×F)-F\cdot(∇×G)$$ $$=G_1 (∂F_3)/∂y-G_1 (∂F_2)/∂z+G_2 (∂F_1)/∂z-G_2 (∂F_3)/∂x+G_3 (∂F_2)/∂x-G_3 (∂F_1)/∂y-F_1 (∂G_3)/∂y+F_1 (∂G_2)/∂z-F_2 (∂G_1)/∂z+F_2 (∂G_3)/∂x-F_3 (∂G_2)/∂x+F_3 (∂G_1)/∂y$$ and $$-G_2 (∂F_3)/∂x+G_3 (∂F_2)/∂x+F_2 (∂G_3)/∂x-F_3 (∂G_2)/∂x$$ $$=(G_3 ∂F_2-G_2 ∂F_3+F_2 ∂G_3-F_3 ∂G_2)/∂x$$
We can use multi-index notation. Note \begin{equation} \begin{aligned} \nabla\cdot (F\times G) & =\nabla_\mu(F\times G)^\mu=\partial_\mu\left(\epsilon^{\mu l k}F_lG_k\right)= \epsilon^{\mu l k}\partial_\mu(F_l G_k)= G_k \epsilon^{\mu l k}\partial_\mu F_l +F_l \epsilon^{\mu l k}\partial_\mu G_k \\ &= G_k \epsilon^{k\mu l}\partial_\mu F_l -F_l \epsilon^{l\mu k}\partial_\mu G_k= G \cdot (\nabla \times F)-F \cdot (\nabla\times G) \end{aligned} \end{equation} where we used the commutation properties of the symbol $\epsilon^{\mu l k}$.