triple vector product: vector vs gradient I think there's a simple explanation for this, but could not find one from a few online searches. The triple vector product and the curl of $\mathbf{A}\times \mathbf{B}$ have very similar forms, however there are additional terms in the differentiation case:
$
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \bullet \mathbf{C}) - \mathbf{C} (\mathbf{A} \bullet \mathbf{B}) \\
  \nabla \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\nabla \bullet \mathbf{C}) - \mathbf{C} (\nabla \bullet \mathbf{B}) + (\mathbf{C} \bullet \nabla)\mathbf{B} - (\mathbf{B} \bullet \nabla)\mathbf{C}
$
Can someone explain why? I'm familiar with Einstein notation if that helps. Thank you for your help!
 A: I thought it might be useful to present way forward that I prefer.  Here, we begin by introducing notation.  
Let $(x_1,x_2,x_3)$ be Cartesian coordinates.  We designate by $\hat x_i$, a unit vector along the $x_i$ axis and by $\partial_i$ the partial derivative with respect to $x_i$.  
Then, using the convention of summing over repeated indices, the $i$'th component of the vector $\nabla \times (\vec B\times \vec C)$ can be written as 
$$\begin{align}
\hat x_i\cdot \left(\nabla \times (\vec B\times \vec C)\right)&=\hat x_i\cdot \left(\hat x_j\times (\hat x_k\times \hat x_\ell)\right)\partial_j(B_kC_\ell)\tag 1\\\\
&=\left(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}\right)\left(B_k \partial_j(C_\ell)+C_\ell \partial_j(B_k)\right)\tag 2\\\\
&=B_i\partial_j(C_j)-B_j\partial_j(C_i)+C_j\partial_j(B_i)-C_i\partial_j(Bj)\tag 3\\\\
&=B_i(\nabla \cdot \vec C)-(\vec B\cdot \nabla)C_i+(\vec C\cdot \nabla)B_i-C_i(\nabla \cdot \vec B) \tag 4
\end{align}$$
In going from $(1)$ to $(2)$, we used the vector triple product rule (presumed established) while in going from $(2)$ to $(3)$ we used the sifting property of the Kronecker Delta.  
Since $(4)$ is true for all $i$, then upon multiplying by $\hat x_i$ and summing over $i$, we find the coveted identity
$$\nabla \times (\vec B\times \vec C)=\vec B(\nabla \cdot \vec C)-(\vec B\cdot \nabla)\vec C+(\vec C\cdot \nabla)\vec B-\vec C(\nabla \cdot \vec B)$$
A: As stated above, the first expression given is simply product of vectors, which can be expressed in terms of the dot product. The second involves differentiation, acting on a product. The product rule for vector differentiation will inevitably lead to the extra terms.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mrm}[1]{\,\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\epsilon_{abc}}$ is the
  Levi-Civita Symbol or/and the Levi-Civita Tensor.


\begin{align}
\color{#f00}{\vec{A}\times\pars{\vec{B}\times\vec{C}}} & =
\sum_{i}\hat{e}_{i}\bracks{\vec{A}\times\pars{\vec{B}\times\vec{C}}}_{i} =
\sum_{i}\hat{e}_{i}\sum_{jk}\epsilon_{ijk}\,A_{j}\pars{\vec{B}\times\vec{C}}_{k}
\\[5mm] & =
\sum_{ijk}\hat{e}_{i}\,\epsilon_{ijk}\,A_{j}
\sum_{\ell m}\epsilon_{k\ell m}\,\,B_{\ell}\,C_{m} =
\sum_{ij\ell m}\hat{e}_{i}\,A_{j}\,B_{\ell}\,C_{m}\
\underbrace{\sum_{k}\epsilon_{ijk}\,\,\epsilon_{\ell mk}}
_{\ds{\delta_{i\ell}\,\delta_{jm} - \delta_{im}\,\delta_{j\ell}}}
\\[5mm] & =
\sum_{ij}\hat{e}_{i}A_{j}B_{i}C_{j} - \sum_{ij}\hat{e}_{i}A_{j}B_{j}C_{i}
\\[5mm] & =
\pars{\sum_{i}\hat{e}_{i}B_{i}}\pars{\sum_{j}A_{j}C_{j}} -
\pars{\sum_{i}\hat{e}_{i}C_{i}}\pars{\sum_{j}A_{j}B_{j}}
\\[5mm] & =
\color{#f00}{\vec{B}\pars{\vec{A}\cdot\vec{C}} - \vec{C}\pars{\vec{A}\cdot\vec{B}}}
\end{align}


The other one is somehow similar. You can take the opportunity to 'play' with the Levi-Civita $\ds{\epsilon_{abc}}$ Symbol.

