Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 12 / i / Solution

In components the scalar triple product is $A_i{ϵ}_{ijk}{B}_j{C}_k$. A cyclic permutation has positive sign of a permutation, so ${ϵ}_{ijk}={ϵ}_{kij}$. Relabelling the summed indices therefore giveswhich proves $A\cdot (B\times C)=(A\times B)\cdot C$. Equivalently both expressions are the same three-column determinant.