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!

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$\ds{\epsilon_{abc}}$ 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.

