TITLE: Show $\nabla\cdot\left(\mathbf{F}\times\mathbf{G}\right)=\mathbf{G}\cdot(\nabla\times\mathbf{F})-\mathbf{F}\cdot(\nabla\times\mathbf{G})$
QUESTION [9 upvotes]: Question as follows.
Suppose that $\mathbf{F}$,$\mathbf{G}:\mathbb{R^3}\rightarrow\mathbb{R^3}$ and $\phi:\mathbb{R^3}\rightarrow\mathbb{R}$ are smooth. Show using the summation convention that $$\nabla\cdot\left(\mathbf{F}\times\mathbf{G}\right)=\mathbf{G}\cdot(\nabla\times\mathbf{F})-\mathbf{F}\cdot(\nabla\times\mathbf{G}).$$
So far I have $$\mathrm{LHS}=\partial_i\mathbf{e}_i\cdot\left(F_j\mathbf{e}_j\times G_k\mathbf{e}_k\right)=\partial_i\mathbf{e}_i\cdot(F_j G_k \epsilon_{jki}\mathbf{e}_i)=\partial_iF_jG_k\epsilon_{ijk}.$$
I'm under the impression that this should be $0$ as expanding the $\mathrm{RHS}$ gives $2\partial_iF_jG_k\epsilon_{ijk}$. Is this true and is it true because for $\partial_iF_jG_k\ne0$ iff $i=j$ or $i=k$ but if $i=j$ or $i=k$ then $\epsilon_{ijk}=0$?

REPLY [2 votes]: The same calculation that Kaster, fully expanded:
\begin{align*}
\nabla\cdot(\mathbf{F}\times\mathbf{G})
&= \partial_1(F_2\,G_3 - F_3\,G_2) + \partial_2(F_3\,G_1 - F_1\,G_3) + \partial_
3(F_1\,G_2 - F_2\,G_1)\\
&= (\partial_1 F_2)\,G_3 - (\partial_1 F_3)\,G_2
 + (\partial_2 F_3)\,G_1 - (\partial_2 F_1)\,G_3
 + (\partial_3 F_1)\,G_2 - (\partial_3 F_2)\,G_1\\
&\qquad+ F_2\,(\partial_1 G_3) - F_3\,(\partial_1 G_2)
 + F_3\,(\partial_2 G_1) - F_1\,(\partial_2 G_3)
 + F_1\,(\partial_3 G_2) - F_2\,(\partial_3 G_1)\\
&= G_1\,(\partial_2 F_3 - \partial_3 F_2) + G_2\,(\partial_3 F_1 - \partial_1 F_
3) + G_3\,(\partial_1 F_2 - \partial_2 F_1)\\
&\qquad- F_1\,(\partial_2 G_3 - \partial_3 G_2) - F_2\,(\partial_3 G_1 - \partial_1 G_3) - F_3\,(\partial_1 G_2 - \partial_2 G_1)\\
&= \mathbf{G}\cdot(\nabla\times\mathbf{F}) - \mathbf{F}\cdot(\nabla\times\mathbf{G}).
\end{align*}