Proving that $X\times (Y\times Z)+Y\times (Z\times X)+Z\times (X\times Y)=0$ for all $X, Y, Z\in \mathbb{R^3}$

Question

I am looking for a proof that $$X\times (Y\times Z)+Y\times (Z\times X)+Z\times (X\times Y)=0 \qquad \textrm{for all } X, Y, Z\in \mathbb{R^3} .$$

I know that as the left-hand side is a summation of all the even permutations, so it should be zero. However, I am looking for some clearer and understandable proof. Any comments?

Answer 1

I suspect this question is a duplicate, but I cannot find another instance of it. In any case:

Hint Rewrite the left-hand side of the identity using the vector triple product identity that expresses the iterated cross product in terms of the dot product: $${\bf x} \times ({\bf y} \times {\bf z}) = ({\bf x} \cdot {\bf z}) {\bf y} - ({\bf x} \cdot {\bf y}) {\bf z} .$$

Answer 2

From the fact $X\times Y$ is alternating (i.e. $Y\times X=-X\times Y$), deduce

$$ (X,Y,Z):=X\times(Y\times Z)+Y\times(Z\times X)+Z\times(X\times Y) $$

is alternating. Moreover, $\times$ is rotationally invariant, i.e. $(RX)\times(RY)=R(X\times Y)$ for all rotation matrices $R$ (this is because $X\times Y$ is uniquely determined by $X$ and $Y$ and the geometry of space), so the multilinear form $(X,Y,Z)$ must be as well. Thus, WLOG we can consider $X=e_1$, then from the fact it's alternating and linear WLOG $Y=e_2$ and $Z=e_3$ and then it's a single calculation.