If 1/a+1/b+1/c=1/(a+b+c) where a+b+c not equal to zero and abc not equal to zero.what is the value of (a+b)(b+c)(c+a)

1)=0 2)>0 3)<0 4) cannot b determined

• Answer is 1)=0
  Expalanation:-
  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)}/(a+b+c)*c .....................(a+b+c)& abc are non-zero
  (a+b)(a+b+c)c = ab{c-(a+b+c)
  (a+b)(a+b+c)c=-ab(a+b)
  (a+b)(a+b+c)c + ab(a+b)=0
  (a+b){(a+b+c)c+ab}=0
  (a+b){ac+bc+c2+ab}=0
  (a+b){a(c+b)+c(a+b)}=0
  (a+b)(b+c)(a+c)=0
• 9 years agoHelpfull: Yes(32) No(1)
• 1/a+1/b+1/c=1/(a+b+c)
  or, (ab+bc+ca)/abc= 1/(a+b+c)
  or, (a+b+c)*(ab+bc+ca)=abc
  or, a^2b+a^2c+ab^2+ac^2+b^2c+bc^2+3abc=abc
  or, a^2b+a^2c+ab^2+ac^2+b^2c+bc^2+2abc=0
  or, a^2b+a^2c+ab^2+ac^2+b^2c+bc^2+abc+abc=0
  or, (a+b)(b+c)(c+a)=0
• 9 years agoHelpfull: Yes(3) No(0)
• Ans:--- 1)0
• 9 years agoHelpfull: Yes(0) No(0)
• 0 will be the value.
• 9 years agoHelpfull: Yes(0) No(0)
• 0 is write answer by hit n trial method
  
• 9 years agoHelpfull: Yes(0) No(0)
• take a=1 b=1 c=-1 ;)
  
  Answer 0
• 9 years agoHelpfull: Yes(0) No(0)