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Question
Using properties of scalar triple product, prove that `[(bara + barb, barb + barc, barc + bara)] = 2[(bara, barb, barc)]`.
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Solution
L.H.S = `[(bara + barb, barb + barc, barc + bara)]`
= `(bara + barb) . [(barb + barc) xx (barc + bara)]`
= `(bara + barb) . [barb xx barc + barb xx bara + barc xx barc + barc xx bara]`
= `(bara + barb) . [barb xx barc + barb xx bara + barc xx bara] ...[∵ barc xx barc = bar0]`
= `bara . [(barb xx barc) + (barb xx bara) + (barc xx bara)] + barb . [(barb xx barc) + (barb xx bara) + (barc xx bara)]`
= `bara . (barb xx barc) + bara . (barb xx bara) + bara . (barc xx bara) + barb . (barb xx barc) + barb(barb xx bara) + barb(barc xx bara)`
= `[bara barb barc] + [bara barb bara] + [bara barc bara] + [barb barb barc] + [barb barb bara] + [barb barc bara]`
= `[bara barb barc] + 0 + 0 + 0 + 0 + [bara barb barc]`
= `2[bara barb barc]`
= R.H.S
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