If 1/a+1/b+1/c=1/(a+b+c) where a+b+c not equals=0, abc not equals=0, what is the value of (a+b) (b+c) (A+c)...??

a. Equals 0
b.Greater than 0
c.Less than 0
d. Can't be determined

• let assume the value of a,b &c that will satisfy the equation.
  so a=1,b=1,c= -1 or a=1,b= -1 &c=1 or a= -1,b=1,c=1
  here all three cases satisfy the equation
  assume any one a=1,b=1,c= -1
  hence (a+b)(b+c)(c+a)=(1+1)(1-1)(-1+1)=0
• 7 years agoHelpfull: Yes(8) 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
• 7 years agoHelpfull: Yes(7) No(0)
• 1/a+1/b+1/c = 1/(a+b+c)
  take lcm:
  (bc+ca+ab)/abc = 1/(a+b+c)
  
  abc+ ca^2+a^2b+b^2c+cab+ab^2+bc^2+c^2a+abc= abc
  
  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
• 7 years agoHelpfull: Yes(0) No(0)