. If 1/a + 1/b + 1/c = 1/(a+b+c) where a + b + c != 0 , abc != 0, what is the value of (a+b)(b+c)(c+a)?

A. Equals 0

B. Greater than 0

C. Less than 0

D. Cannot be determined

• first solve the eq-
  1/a+1/b+1/c=1/(a+b+c)
  =>(bc+ac+ab)/abc=1/(a+b+c)
  =>(a+b+c)(bc+ac+ab)=abc
  =>abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2+abc=abc
  =>2abc+a^2c+a^2b+b^2c+ab^2+bc^2+ac^2=0 ----eq (1)
  
  Now solve the eq-
  (a+b)(b+c)(c+a)
  =>(ab+ac+b^2+bc)(c+a)
  =>abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2+abc
  =>2abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2 ---eq(2)
  Now compare eq(1) and eq(2)
  Since both are equal ao we conclude that
  (a+b)(b+c)(c+a)=0
  So the ans is optn (a)
  
• 10 years agoHelpfull: Yes(30) No(1)
• put a=1,b=-1,c=1 then 1/a+1/b+1/c=1 and 1/(a+b+c)=1
  so (a+b)(b+c)(c+a)=0 ans
• 10 years agoHelpfull: Yes(7) No(8)
• Since abc != 0 and a+b+c !=0 ; when we take modify the term 1/a + 1/b + 1/c = 1/(a+b+c) , we can simply multiply (a+b+c) and abc on numberator. That will yield us, (a+b+c)(bc+ca+ab) = abc or 2abc + a^2(b+c) + b^2(a+c) + c^2(a+b) = 0;
  
  when we simplify (a+b)(b+c)(c+a), we get 2abc + a^2(b+c) + b^2(c+a) +c^2(a+b) . So, the answer is simply 0.
• 10 years agoHelpfull: Yes(2) No(2)
• equals to zero.
  
  
  first solve the eq-
  1/a+1/b+1/c=1/(a+b+c)
  =>(bc+ac+ab)/abc=1/(a+b+c)
  =>(a+b+c)(bc+ac+ab)=abc
  =>abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2+abc=abc
  =>2abc+a^2c+a^2b+b^2c+ab^2+bc^2+ac^2=0 ----eq (1)
  
  Now solve the eq-
  (a+b)(b+c)(c+a)
  =>(ab+ac+b^2+bc)(c+a)
  =>abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2+abc
  =>2abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2 ---eq(2)
  Now compare eq(1) and eq(2)
  Since both are equal ao we conclude that
  (a+b)(b+c)(c+a)=0
  So the ans is optn (a)
• 10 years agoHelpfull: Yes(2) No(0)
• let S=(a+b)(b+c)(c+a)
  
  putting different values of a,b,c in 'S' we will get different values of it.
  like putting a=3,b=-1,c=-1 we get -ve value of S.
  
  if we put a=1,b=-1,c=1 we get S=0
  
  we will get different values of 'S' changing the values of a,b,c satisfying the equations a+b+c!=0 and abc!=0
  
  so answer may be option D.
• 10 years agoHelpfull: Yes(1) No(2)
• Ans will be B
• 10 years agoHelpfull: Yes(0) No(1)
• a equals 0
  
• 10 years agoHelpfull: Yes(0) No(1)
• b) greater then 0
• 10 years agoHelpfull: Yes(0) No(1)
• c less than 0
  
• 10 years agoHelpfull: Yes(0) No(1)
• A.equals 0
• 10 years agoHelpfull: Yes(0) No(2)
• equal to 0...assume a=-1,b=1,c=1
• 9 years agoHelpfull: Yes(0) No(0)
• 1/a + 1/b + 1/c =1/ (a+b+c)
  => 1/a + 1/b =1/ (a+b+c) - 1/c
  => (b+a)/ab = (c - (a+b+c)) / c(a+b+c)
  
  Cross multiplying (Since, a,b,c !=0 and a+b+c != 0)
  
  => c(a+b+c)(b+a) = - ab(a+b)
  => (a+b)( ca +cb +cc + ab ) = 0
  => (a+b)( ca + cc + cb + ab) = 0
  => (a+b)(b+c)(c+a)=0
• 9 years agoHelpfull: Yes(0) No(0)