Prove [a+b b+c c+a] = 2[a b c]

[a + b b + c c + a] = (a + b) · [(b + c) × (c + a)].

Expand the cross product

(b + c) × (c + a) = b × c + b × a + c × c + c × a
                  = b × c + b × a + 0 + c × a

Dot with (a + b)

(a + b) · (b × c) = a · (b × c) + b · (b × c) = [a b c] + 0
(a + b) · (b × a) = a · (b × a) + b · (b × a) = 0 + 0
(a + b) · (c × a) = a · (c × a) + b · (c × a) = 0 + [b c a] = [a b c]

(a triple product with a repeated vector is zero, and [b c a] = [a b c] by cyclic permutation.)

Add

[a + b b + c c + a] = [a b c] + 0 + [a b c].

Hence [a + b b + c c + a] = 2[a b c].