The vector triple product, also known as the BAC-CAB identity, is used to calculate the cross product when three vectors are involved. First, a few words about notation:

Let **A**, **B** and **C** be vectors in **R**^{3}.

**A**×(**B**×**C**)=**B**(**A**·**C**)-**C**(**A**·**B**)

This is much easier to prove if we pick our coordinate system such that **B** lies along the x-axis and **C** lies in the x-y plane. Since we are dealing with proper vectors, if we show this is true in one coordinate system, it must be true in all coordinate systems.

**B**=(B_{x}, 0, 0), **C**=(C_{x}, C_{y}, 0), **A**=(A_{x}, A_{y}, A_{z})

**B**×**C**=(0, 0, B_{x}C_{y})

**A**×(**B**×**C**)=(A_{x}, A_{y}, A_{z})×(0, 0, B_{x}C_{y})

=(A_{y}B_{x}C_{y}, -A_{x}B_{x}C_{y}, 0)

**B**(**A**·**C**)-**C**(**A**·**B**)=[(A_{x}C_{x}+A_{y}C_{y})B_{x}, 0, 0] - [A_{x}B_{x}C_{x}, A_{x}B_{x}C_{y}, 0]

=(A_{y}B_{x}C_{y}, -A_{x}B_{x}C_{y}, 0)

**A**×(**B**×**C**)=**B**(**A**·**C**)-**C**(**A**·**B**) QED

A few tidbits:

(**A**×**B**)×**C**= -**C**×(**A**×**B**)

= **B**(**A**·**C**)-
**A**(**B**·**C**) *Parentheses are important!*

**A**×(**B**×**C**)+**B**×(**C**×**A**)+**C**×(**A**×**B**)=0

All higher vector products (products involving more than three vectors) can be reduced to expressions containing no more than one cross product per term, often due to the BAC-CAB identity.

Sources:

Bressoud, D. *Second Year Calculus* © 1991

http://farside.ph.utexas.edu/teaching/em1/lectures/node11.html