Skew symmetric matrix of vector triple product

Question

Do someone know how to describe the skew symmetric matrix form of $B \times(C\times D)$ vector triple product? I know that $[A\times B]_\times = [A]_\times[B]_\times - [B]_\times[A]_\times$ as described here. I am asking because I have the following: $$ A\times [B \times (C \times D)] $$ which I have rewritten in the matrix form: $$ [A]_\times[B]_\times[C]_\times D $$ And I need to swap positions and have $[B]_\times$ in the first position, for example $[B]_\times[C]_\times[D]_\times A$, without changing the result. I was thinking of doing so: $$ A\times [B \times (C \times D)] = -[B \times (C \times D)]\times A $$ But I need to put that in a matrix form using the skew symmetric matrix, but I don't know how. If someone have a suggestion, I would appreciate the help. Thank you.

Answer 1

If you want to rewrite $A\times(B\times(C\times D))$ as $B\times V(A,C,D)$ for some function $V$, it's impossible.

Suppose the contrary. Then $B\cdot(A\times(B\times(C\times D)))=0$ for all $A,B,C,D$. In particular, we would have $$ 1=\mathbf i\cdot(\mathbf j\times(\mathbf i\times(\mathbf k\times\mathbf i)))=0, $$ which is a contradiction.

Anyway, by Jacobi's identity $B\times(C\times D)+C\times(D\times B)+D\times(B\times C)=0$, you may express $B\times(C\times D)$ in terms of $C\times(D\times B)$ and $D\times(B\times C)$.