(b×c)×(c×a)=[a b c]c and deduction

Part 1: (b × c) × (c × a) = [a b c] c

Use P × (Q × R) = Q(P · R) − R(P · Q) with P = b × c, Q = c, R = a:

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

• (b × c) · a = [a b c] (the cyclic form of a · (b × c)).
• (b × c) · c = 0, because b × c is perpendicular to c.

So (b × c) × (c × a) = [a b c] c.

Part 2: Deduce [b×c c×a a×b] = [a b c]²

By definition [u v w] = (u × v) · w. With u = b × c, v = c × a, w = a × b:

[b×c  c×a  a×b] = [(b × c) × (c × a)] · (a × b)
                = [a b c] c · (a × b)       (using Part 1)
                = [a b c] [c a b]
                = [a b c] [a b c]            (cyclic permutation)

Hence [b×c c×a a×b] = [a b c]².

Check with a = i, b = j, c = k: (j×k)×(k×i) = i × j = k = [i j k] k ✓.