# 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 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 $\epsilon_{ijk}=0$?

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Remember that the left-hand side needs the product rule - the partial derivative $\partial_i$ does not commute with $F_j$ and $G_k$ –  Michael Jun 23 '13 at 6:48
So I don't even need the condition $i=k$? Saying that $i=j\Rightarrow \epsilon_{ijk}=0$ is enough? –  jamesh625 Jun 23 '13 at 7:04
No, $\partial_i(F_jG_k)=F_{j,i}G_k+F_jG_{k,i}$, where the comma means the partial derivative. –  Michael Jun 23 '13 at 7:11
$$\nabla \cdot \left ( \mathbf F \times \mathbf G\right ) = \left( \mathbf F \times \mathbf G\right )_{i,i} = \left ( e_{ijk} F_j G_k\right )_{,i} = e_{ijk}F_{j,i}G_k + e_{ijk}F_j G_{k,i} = \\ = G_k e_{kij}F_{j,i} + F_j e_{jki} G_{k,i} = G_k e_{kij}F_{j,i} - F_j e_{jik} G_{k,i} = \\ = G_k \left (\nabla \times \mathbf F \right )_k - F_j \left( \nabla \times \mathbf G\right )_j = \mathbf G \cdot \left (\nabla \times \mathbf F \right ) - \mathbf F \cdot \left( \nabla \times \mathbf G\right )$$