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Question

Prove that: a(b+c)×(a+2b+3c)=[abc]

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Solution

L.H.S.=a(b+c)×(a+2b+3c)
=a(b×b×2b+b×3c+c×a+c×2b+c×3c)
=a(b×a+3b×c+c×a+2c×b)
=a(b×a)+a(3b×c)+a(c×a)+c(2c×b)
=0+3[a(b×c)+0+2(a(c×b)]
=3[abc]+2[abc]
=3[abc]2[abc]
=[abc]=R.H.S.

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