Proving a vector identity ∇ · (A × B) this is my first post here :)
Nice forum by the look of things.
Anyway my teacher showed me this I will attach the link here:
It is page 2 from this link: http://www2.ph.ed.ac.uk/~mevans/mp2h/VTF/lecture15.pdf
At 15.3. the pdf discusses: Products of Two Vector Fields:
When the article discusses the proof of (6) is that the full proof of proving the identity?
Essentially it is this:
I would like to prove: $$\nabla \cdot (A \times B) = B \cdot (\nabla \times A) − A \cdot (\nabla \times B)$$
And the pdf shows this: 
\begin{eqnarray*}
\nabla \cdot (A \times B) &=& \\
&=& \frac{\partial}{\partial x_i} e_{ijk} A_j B_k \\
&=& e_{ijk}\frac{\partial A_j}{\partial x_i}B_k + e_{ijk} A_j \frac{\partial B_k}{\partial x_i} \\
&=& B_k e_{kij} \frac{\partial A_j}{\partial x_i} - A_j e_{jik}\frac{\partial B_k}{\partial x_i}
\end{eqnarray*}
(It is a lot easier on the pdf)
It just seems that the proof unfinished and doesn't show how it gets back to the form of:
$$B \cdot (\nabla \times A) − A \cdot (\nabla \times B)$$
Also if anyone knows can the letters A and B be exchanged by any letters say C and D for example?
Thanks if anyone read this and can help.
-nomad609
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\nabla\cdot\pars{\vec{A}\times\vec{B}} & =
\sum_{i}\partiald{\pars{\vec{A}\times\vec{B}}_{i}}{x_{i}} =
\sum_{i}\partiald{}{x_{i}}\sum_{jk}\epsilon_{ijk}A_{j}B_{k} =
\sum_{ijk}\epsilon_{ijk}\,\partiald{A_{j}}{x_{i}}\,B_{k} +
\sum_{ijk}\epsilon_{ijk}\, A_{j}\,\partiald{B_{k}}{x_{i}}
\\[5mm] & =
\sum_{k}\pars{\sum_{ij}\epsilon_{kij}\,\partiald{}{x_{i}}\,A_{j}}B_{k} -
\sum_{j}\pars{\sum_{ik}\epsilon_{jik}\,\partiald{}{x_{i}}\,B_{k}}A_{j}
\\[5mm] & =
\sum_{k}\pars{\nabla\times\vec{A}}_{k}B_{k} -
\sum_{j}\pars{\nabla\times\vec{B}}_{j}A_{j} =
\bbx{\vec{B}\cdot\nabla\times\vec{A} - \vec{A}\cdot\nabla\times\vec{B}}
\end{align}
A: 
I thought it might be useful to remark that the Levi-Civita symbol, $\epsilon_{ijk}$ can be written alternatively as 
$$\bbox[5px,border:2px solid #C0A000]{\epsilon_{ijk}=\hat x_i\cdot (\hat x_j\times\hat x_k) }\tag 1$$
in terms of the Cartesian unit vectors $\hat x_i$, $i=1,2,3$.  

Then, using $(1)$ we can write
$$\begin{align}
\epsilon_{ijk}\left(B_k\frac{\partial A_j}{\partial x_j}\right)&=\hat x_i\cdot (\hat x_j\times\hat x_k)\left(B_k\frac{\partial A_j}{\partial x_i}\right)\\\\
&=\hat x_k\cdot (\hat x_i\times\hat x_j)\left(B_k\frac{\partial A_j}{\partial x_i}\right)\\\\
&=(\hat x_kB_k)\cdot \left(\hat x_i\frac{\partial }{\partial x_x}\right)\times (\hat x_jA_j)\\\\
&=\vec B\cdot \nabla \times \vec A
\end{align}$$

Evidently, this development differs from one that uses the Levi-Civita symbol by notation only.  It does assign, however, to $\epsilon_{ijk}$ a meaning as the scalar triple product.  And in doing so, this approach affords more transparency on the interaction among the indices of $\epsilon_{ijk}$ as unit vectors and the associated Cartesian components of both the "Del" operator, $\nabla$, and the vectors $\vec A$ and $\vec B$ on which it operates.

