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(C) Using the vector triple product identity $(p 	imes q) 	imes r = (p \cdot r)q - (q \cdot r)p$,let $p = b$,$q = c$,and $r = (c 	imes a)$.Then,$(b

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$[a, b, c] c$

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(C) Using the vector triple product identity $(p 	imes q) 	imes r = (p \cdot r)q - (q \cdot r)p$,let $p = b$,$q = c$,and $r = (c 	imes a)$.Then,$(b 	imes c) 	imes (c 	imes a) = (b \cdot (c 	imes a))c - (c \cdot (c 	imes a))a$.Since $c \cdot (c 	imes a) = 0$ (as the vector $c 	imes a$ is perpendicular to $c$),the expression simplifies to $[b, c, a]c$.Using the cyclic property of the scalar triple product,$[b, c, a] = [a, b, c]$.Therefore,the result is $[a, b, c]c$.

$(b \times c) \times (c \times a) = $

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