TITLE: Cross-product identity
QUESTION [5 upvotes]: This page of vector identities lists the following (among many other identities):
$$
(\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}))\,\mathbf{D}= (\mathbf{A}\cdot\mathbf{D} )\left(\mathbf{B}\times\mathbf{C}\right)+\left(\mathbf{B}\cdot\mathbf{D}\right)\left(\mathbf{C}\times\mathbf{A}\right)+\left(\mathbf{C}\cdot\mathbf{D}\right)\left(\mathbf{A}\times\mathbf{B}\right)
$$
which is presumably supposed to hold for vectors $\mathbf{A,B,C,D} \in \Bbb R^3$. Unlike the other identities, this one is given without justification or citation.  With this in mind, my questions are:

Is the identity true? (proven in answers below)
Is the identity well-known? Is there a citation that can be used here?
How can we prove it?

Some answers have been given, but alternate approaches would be interesting to see.
Thank you for your consideration.

Quick thoughts on the problem:

$\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})$ is a scalar-triple product and can be rewritten as
$$ 
\det \pmatrix{\mathbf{A}& \mathbf{B} & \mathbf{C}}
$$
I have a hunch Cauchy-Binet can be applied here somehow
This amounts to a statement about the map
$$
D \mapsto [(A \times B)(C\cdot D) + (B \times C)(A\cdot D) + (C \times A)(B\cdot D)]
$$
A proof in Levi-Cevita notation might be quick.

REPLY [2 votes]: Since the two sides of the equation are linear in each factor, we may reduce $A, B, C$ to basis vectors. Further, if two of $A, B, C$ are equal, then both sides are $0$. Thus we may assume $A=e_i,\,B=e_j,\,C=e_k$ with $i,j,k$ mutually distinct.
In this case $B\times C$ is a scalar multiple of $A$, so $B\times C=((B\times C)\cdot A)A$.
Denote the matrix $(B \times C)A^T + (C \times A)B^T + (A \times B)C^T$ as $E$. Then the $i$-th column of $E$ is (matrix multiplies on a column):
$$E\cdot A=B\times C=((B\times C)\cdot A)A=\det \pmatrix{\mathbf{A}& \mathbf{B} & \mathbf{C}}A.$$
Here note that $e_i^T e_j=\begin{cases}1,&i=j\\0,&i\ne j\end{cases}$.
Similarly, we can show that the $j$-th and the $k$-the columns of $E$ are the same as those of $\det \pmatrix{\mathbf{A}& \mathbf{B} & \mathbf{C}}I$.
Therefore $$E=\det \pmatrix{\mathbf{A}& \mathbf{B} & \mathbf{C}}I,$$
which is what we want to prove.
Remark:
In essence this is a proof using Levi-Civita notation, since $e_i\times e_j=\epsilon_{ijk}e_k$.
Edit:
I shall show the reduction step here for the rigor.
Write $A=\sum a_ie_i$, $B=\sum b_ie_i$, and $C=\sum c_ie_i$.
Then 
\begin{align*}
  (B \times C)A^T + (C \times A)B^T + (A \times B)C^T
  &=(\sum b_ie_i\times{\sum c_ie_i}) (\sum a_ie_i)^T + \cdots\\
  &=\sum_{i,j,k}(b_ic_ja_k)\left((e_i\times e_j)e_k^T+(e_j\times e_k)e_i^T+(e_k\times e_i)e_j^T\right).
\end{align*}
And clearly $\det\pmatrix{A&B&C}=\sum_{i,j,k}b_ic_ja_k\det\pmatrix{e_k&e_i&e_j}$.
So if we can prove the equation ofr basis vectors, then the equation holds.

Hope this helps.