Identity with four vectors (a×b)×(c×d)+...

We use (P × Q) × (R × S) = [P R S] Q − [Q R S] P, which follows from (P × Q) × Y = Q(P · Y) − P(Q · Y) with Y = R × S.

Expand each term

(a×b)×(c×d) = [a c d] b − [b c d] a
(a×c)×(d×b) = [a d b] c − [c d b] a
(a×d)×(b×c) = [a b c] d − [d b c] a

In the second and third terms, the triple products multiplying a are [c d b] = [b c d] and [d b c] = [b c d] (cyclic permutation).

Add

The a-terms give −3[b c d] a. With the remaining terms:

Sum = −3[b c d] a + [a c d] b + [a d b] c + [a b c] d

Now [a d b] = −[a b d], so the last three terms are [a c d] b − [a b d] c + [a b c] d.

Key relation for any four vectors

Take b, c, d non-coplanar. Then a = x b + y c + z d. Taking the dot product with c × d, d × b and b × c in turn gives

x = [a c d]/[b c d], y = [b a d]/[b c d], z = [b c a]/[b c d],

so [b c d] a = [a c d] b + [b a d] c + [b c a] d = [a c d] b − [a b d] c + [a b c] d.

Hence [a c d] b − [a b d] c + [a b c] d = [b c d] a. (Both sides are polynomials in the components, so the relation also holds when b, c, d are coplanar, by continuity.)

Finish

Sum = −3[b c d] a + [b c d] a = −2[b c d] a, as required.