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# If a, b, c are distinct positive numbers, then the expression (b + c - a)(c + a - b)(a + b - c) - abc is A. positive B. negative C. both negative and positive D. none of the above

SHAIK AASIF AHAMED
7 years ago
Without loss of generality we can assume thata>b>c
Let
(−a+b+c)(a−b+c)(a+b−c)=S
⇒S=(−a+b+c){a−(b−c)}{a+(b−c)}
⇒S=(−a+b+c){a2−(b−c)2}
⇒S=(−a+b+c){a2−b2−c2+2bc}
⇒S=−(a3+b3+c3)−2abc+b2c+bc2+ab2+a2b+ac2+a2c
⇒abc−S=(a3+b3+c3)+3abc−(b2c+bc2+ab2+a2b+ac2+a2c)
⇒abc−S=(a3−a2b)+(b3−b2c)+(c3−c2a)+(abc−bc2)+(abc−ab2)+(abc−a2c)
⇒abc−S=a2(a−b)+b2(b−c)+c2(c−a)+bc(a−c)+ab(c−b)+ac(b−a)
⇒abc−S=a(a−b)(a−c)+b(b−c)(b−a)+c(c−a)(c−b)
⇒abc−S=(a−b){a(a−c)−b(b−c)}+c(c−a)(c−b)
⇒abc−S=(a−b)2{a2−b2+c(b−a)}+c(c−a)(c−b)
⇒abc−S=(a−b)2{a+b−c}+c(c−a)(c−b)
Now(c−a)<0and(c−b)<0
⇒c(c−a)(c−b)>0
and
(a−b)2(a+b−c)>0
This shows
abc−S>0
But in our problem we are asked about S-abc which is always less than 0
so S-abc<0
Hence B is the correct answer
Thanks & Regards
Jitender Singh
IIT Delhi