1/a+1/b+1/c=1/(a+b+c) find (a+b)(b+c)(c+a)

• 1/a+1/b+1/c=1/(a+b+c)
  => (bc+ca+ab)(a+b+c)= abc
  => abc+bc(b+c)+ bca+ca(c+a) +abc+ab(a+b)= abc
  => bc(b+c) + c^2.a + c.a^2 + a^2.b + a.b^2 + 2abc =0
  => bc(b+c) + (c.a^2 + a^2.b) + (c^2.a + a.b^2 + 2abc) =0
  => bc(b+c) + a^2(b+c) + a(b+c)^2 = 0
  => (b+c)[bc+a^2 + ab+ac]=0
  => (b+c)[(bc+ac)+(a^2+ab)]=0
  => (b+c)[c(a+b)+a(a+b)]=0
  => (a+b)(b+c)(c+a)=0
  
  2nd mtd
  
  put a=1,b=-1,c=1 such that a,b,c & (a+b+c) not equals zero & 1/a+1/b+1/c=1/(a+b+c)
  
  so (a+b)(b+c)(c+a)=(1-1)(-1+1)(1+1)=0
  
• 10 years agoHelpfull: Yes(31) No(0)
• 1/a+1/b+1/c = 1/(a+b+c)
  => [(ab+bc+ca)/(abc)]= 1/(a+b+c)
  => a^2.b+a.b^2+b^2.c+b.c^2+a^2.c+a.c^2+3abc=abc
  => (a^2.b+a.b^2+b^2.c+b.c^2+a^2.c+a.c^2+2abc)=0
  since [(a^2.b+a.b^2+b^2.c+b.c^2+a^2.c+a.c^2+2abc)= (a+b)(b+c)(c+a)]
  => (a+b)(b+c)(c+a)=0
  ans is 0
• 10 years agoHelpfull: Yes(6) No(0)
• 1/a+1/b+1/c=1/(a+b+c)
  (ab+bc+ca)/abc=1/(a+b+c)
  (a+b+c)(ab+bc+ca)=abc
  a^2b+ab^2+abc+abc+b^2c+bc^2+a^2c+abc+c^2a=abc
  a^2b+ab^2+b^2c+bc^2+a^2c+c^2a+3abc=abc
  a^2b+ab^2+b^2c+bc^2+a^2c+c^2a+2abc=0
  (a+b)(b+c)(c+a)=0
  
• 10 years agoHelpfull: Yes(4) No(0)
• 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
  abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2=0
  (a+b)(b+c)(c+a)=0
  so the ans will be(=0)
  
• 10 years agoHelpfull: Yes(1) No(0)
• the only possible solution is
  for iit a=1 ,r=2
  for iim a=1 ,d=2
  so
  iit->1+2+4+8=15
  iim->1+3+5+7=16
  total score till first half is
  1+2+1+3=7
  
  
• 10 years agoHelpfull: Yes(0) No(1)
• its 0
  coz
  1/a+1/b+1/c=1/(a+b+c)
  =>abc+a^2c+ba^2+b^2c+abc+b^2a+bc^2+ac^2+abc=abc
  =>(b+a)(b+c)(c+a)=0
• 10 years agoHelpfull: Yes(0) No(0)