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Question

∣ ∣2aa+ba+cb+a2bb+cc+ac+b2c∣ ∣=4(b+c)(c+a)(a+b).

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Solution

LHS=8abc+2(b+c)(c+a)(a+b)+2a(b+c)2+2b(c+a)2+2c(a+b)2

=2(b+c)(c+a)(a+b)+2a(b+c)28abc+2b(c2+2ca+a2)+2c(a2+2ab+b2)

=2(b+c)(c+a)(a+b)+2a(b+c)2+2bc2+2ba2+2ca2+2cb2

=2(b+c)(c+a)(a+b)+2a(b+c)2+2a2(b+c)+2bc(b+c)

=2(b+c)(c+a)(a+b)+2(b+c){a(b+c)+a2+2bc}

=2(b+c)(c+a)(a+b)+2(b+c)(c+a)(a+b)

=4(b+c)(c+a)(a+b)=RHS


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