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!
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 

$$\begin{align}
\hat x_i\cdot \left(\nabla \times (\vec B\times \vec C)\right)&amp;=\hat x_i\cdot \left(\hat x_j\times (\hat x_k\times \hat x_\ell)\right)\partial_j(B_kC_\ell)\tag 1\\\\
&amp;=\left(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}\right)\left(B_k \partial_j(C_\ell)+C_\ell \partial_j(B_k)\right)\tag 2\\\\
&amp;=B_i\partial_j(C_j)-B_j\partial_j(C_i)+C_j\partial_j(B_i)-C_i\partial_j(Bj)\tag 3\\\\
&amp;=B_i(\nabla \cdot \vec C)-(\vec B\cdot \nabla)C_i+(\vec C\cdot \nabla)B_i-C_i(\nabla \cdot \vec B) \tag 4
\end{align}$$

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

$$\nabla \times (\vec B\times \vec C)=\vec B(\nabla \cdot \vec C)-(\vec B\cdot \nabla)\vec C+(\vec C\cdot \nabla)\vec B-\vec C(\nabla \cdot \vec B)$$
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.