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

• 0
  (ab+bc+ca)/abc=1/(a+b+c)
  => abc=a2b+ab2+ac2+a2c+b2c+bc2+3abc
  => a2b+ab2+ac2+a2c+b2c+bc2+2abc=0
  
  (a+b)(b+c)(c+a)=(ab+ac+b2+bc)(c+a)
  =abc+ac2+b2c+bc2+a2c+a2b+b2a+abc
  =0
• 12 years agoHelpfull: Yes(115) No(3)
• a)equals 0
  
  when a= -b
  then LHS of 1/a + 1/b + 1/c=1/(a+b+c)
  1/a + 1/b + 1/c = 1/c = RHS = 1/(a+b+c)
  
  Then a+b=0
  
  hence
  
  (a+b)(b+c)(c+a)=0
  
• 12 years agoHelpfull: Yes(34) No(17)
• 0
  1/a + 1/b + 1/c=1/(a+b+c)
  (ab+bc+ca)/abc = 1/(a+b+c)
  a2c+abc+a2b+abc+b2c+ab2+ac2+bc2+abc=abc
  a2c+a2b+b2c+ab2+ac2+bc2+2abc=0 ------- 1
  
  (a+b)(b+c)(c+a)=a2c+a2b+b2c+ab2+ac2+bc2+2abc=0
  Hence answer should be 0.
  
• 12 years agoHelpfull: Yes(12) No(1)
• hit and trial method.
  1/a + 1/b + 1/c=1/(a+b+c) holds true if a= -c and b can have any other value. take an example of a = 2 , c = -2 and b = 3. so, a = -c i.e. (c+a) = 0 therfore ans is 0.
• 9 years agoHelpfull: Yes(3) No(0)
• 1/a+1/b+1/c=1/(a+b+c)
  by solving this eq we get
  a^2b+a^2c+ab^2+ac^2+bc^2+b^2c+2abc=0.........(1)
  
  now expanding (a+b)(b+c)(c+a)
  wil same as equation (1)
  
  
  so the ans wil "equals 2 zero"
• 10 years agoHelpfull: Yes(2) No(0)
• 1/a +1/b +1/c =1/(a+b+c)
  =>1/a +1/b = (1/(a+b+c)) -(1/c)
  => 1/a +1/b =(c-(a+b+c))/(c(a+b+c))
  => (a+b)/ab =(-(a+b)/c(a+b+c))
  => 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
• 10 years agoHelpfull: Yes(2) No(0)
• greater than 0
• 9 years agoHelpfull: Yes(1) No(0)
• 1/a + 1/b + 1/c=1/(a+b+c) solve this ,
  we will get a2b+ab2+ac2+a2c+b2c+bc2+2abc=0
  (a+b)(b+c)(c+a)=a2b+ab2+ac2+a2c+b2c+bc2+2abc=0
• 8 years agoHelpfull: Yes(1) No(0)
• It is only possible when a,b,c equal to 1. Therefore, the answer is 8
  Therefore ans is
  b) greater than zero
• 8 years agoHelpfull: Yes(1) No(0)
• putting a = -b or b = -c or c = -a gives ans equal to zero
  
• 11 years agoHelpfull: Yes(0) No(2)
• greater than 0
• 10 years agoHelpfull: Yes(0) No(3)
• guys plzz tell how to prepare for the section PS and I m from 2014 batch..plz help..and ty in advance..:)
• 10 years agoHelpfull: Yes(0) No(0)
• 1/a+1/b+1/c= 1/(a+b+c)
  
  => (a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1
  
  subtracting 3 both side,
  => [(a+b+c)/a]-1+[(a+b+c)/b]-1+[(a+b+c)/c]-1=1-3
  
  => (b+c)/a+(a+c)/b+(a+b)/c = -2
  
  adding 2 both side,
  => [(b+c)/a+(a+c)/b+(a+b)/c]+2 = -2+2
  
  => (bc(b+c)+ ac(a+c)+ab(a+b))/(abc) + 2 = 0
  
  => [bc(b+c)+ ac(a+c)+ab(a+b)]+2abc = 0*abc
  
  => bc(b+c)+ ac(a+c)+ab(a+b)+2abc=0
  
  => b^2c+c^2b+a^2c+c^2a+a^2b+b^2a+2abc=0
  
  => a(b^2+ 2bc+c^2) + a^2 b+ a^2c + b^2 c +bc^2 =0
  
  => a(b+c)^2 + a^2(b+c) + bc(b+c) =0
  
  => (b+c) [ a^2 +a(b+c)+ bc]=0
  
  => (b+c) [ a^2 +ab+ac+ bc]=0
  
  => (b+c) [ a(a+b)+c(a+b)]=0
  
  => (b+c)(a +b)(c+a)=0
  
• 8 years agoHelpfull: Yes(0) No(0)
• Why r u expanding this equation just satisfy the values of a b and c let take a=3 b=-1 and c =1 it will satisfy the equation n get the answer 0
• 7 years agoHelpfull: Yes(0) No(0)
• opton b
  less than 0
• 5 years agoHelpfull: Yes(0) No(0)
• 1/a+1/b+1/c=bc+ac+ac/a+b+c
  a+b+c=0
  abc=0
  rhs=1/a+b+c
  =0
• 4 years agoHelpfull: Yes(0) No(0)