Prove that for any three vectors a, b, c $[b + c c + a a + b] = 2 [a b c]$ .

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Solution

$[\to b + \to c \to c + \to a \to a + \to b]$
$(\to b - \to c) {(\to c + \to a) \times (\to a + \to b)}$
$(\to b + \to c) {\to c \times \to a + \to c \times \to b + \to a \times \to a + \to a \times \to b}$
$[b c a] + 0 + 0 + 0 + 0 + 0 + 0 [c a b]$
$[a b c] + [a b c] = 2 [a b c]$

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