The <scalar triple product> is a <determinant>:
$$
(\mathbf a\times\mathbf b)\cdot\mathbf c=\det(\mathbf a,\mathbf b,\mathbf c).
$$
A cyclic permutation of its three columns is an <even permutation>, so it leaves the determinant unchanged. Thus $\det(\mathbf a,\mathbf b,\mathbf c)=\det(\mathbf b,\mathbf c,\mathbf a)=\mathbf a\cdot(\mathbf b\times\mathbf c)$, proving the identity.

Solution

Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 6 / a / i / Solution

 Solution (source code)