Cross-product identity

By formula number 8 in the above link, we may derive from $$A\times((B\times C)\times D)=-A\times (D\times(B\times C))$$ $$\Leftrightarrow (A\cdot D)(B\times C)-(A\cdot(B\times C))D=-A\times((D\cdot C)B-(D\cdot B)C),$$ from which the result follows.

A (reasonably) quick proof of the statement: let $M$ be the matrix whose columns are $A,B,C$. I claim that the adjugate matrix of $M^T$ (i.e. the cofactor matrix of $M$) is given by $$ \operatorname{adj}(M^T) = \pmatrix{B \times C & C \times A & A \times B}. $$ This is simple enough to verify with computation. From there, it follows that $$ \begin{align} (B \times C)A^T + (C \times A)B^T + (A \times B)C^T &= \pmatrix{B \times C & C \times A & A \times B}M^T \\ & = \operatorname{adj}(M^T)M^T = \det(M)I = \det \pmatrix{A & B & C} I. \end{align} $$ Now, take the equation $$ \det \pmatrix{A & B & C} I = (B \times C)A^T + (C \times A)B^T + (A \times B)C^T $$ and multiply (from the right) by the vector $D$. The conclusion follows.

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.

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Hope this helps.