Proof of $BAC-CAB$ identity missing step

Observe that both sides of the equation to be proved are linear in each of the three variables (when the other two are fixed).

Based on this, it's enough to prove the statement for all possible combinations of the standard basis, say.


(This is a long comment.) This is probably not the simplest way to derive the identity, but the identity follows quite quickly from a special case of Cauchy-Binet formula: \begin{aligned} \left(A\times(B\times C)\right)\cdot x &=\det(A,B\times C,x)\quad\text{(defintion of cross product)}\\ &=\det(x,A,B\times C)\\ &=(x\times A)\cdot(B\times C)\quad\text{(defintion of cross product)}\\ &=(x\cdot B)(A\cdot C)-(x\cdot C)(A\cdot B)\quad\text{(Cauchy-Binet formula)}\\ &=\left(B(A\cdot C)-C(A\cdot B)\right)\cdot x.\\ \end{aligned} Since $x$ is arbitrary, the result follows.

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Linear Algebra

Proof Explanation

Vectors

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