3) If 1/a + 1/b + 1/c = 1 / (a + b + c); where a + b + c 1 0; abc 1 0, then what is the value of ( a
+ b ) ( b + c ) ( c + a )?
a) Equal to 0 b) Greater than 0
c) Less than 0 d) Cannot be determined

• a) Equal to 0
  
  1/a+1/b+1/c=1/(a+b+c)
  or (bc+ac+ab)/(abc)=1/(a+b+c)
  or (ab+bc+ac)(a+b+c)=abc
  or abc+(a^2)c+(a^2)b+(b^2)c+abc+(b^2)a+(c^2)b+(c^2)a+abc=abc
  or 2abc+(a^2)c+(a^2)b+(b^2)c+(b^2)a+(c^2)b+(c^2)a=0
  also,
  ( a+ b ) ( b + c ) ( c + a )=2abc+(a^2)c+(a^2)b+(b^2)c+(b^2)a+(c^2)b+(c^2)a
  or ( a+ b ) ( b + c ) ( c + a )=0
• 8 years agoHelpfull: Yes(26) No(3)
• its (a+b+c is) not equal to 0 and (abc) is not equal to 0 in the question
• 8 years agoHelpfull: Yes(3) No(1)
• Answer: a) Equal to 0
  
  Solution:
  [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(2) No(0)
• equal to zero
• 8 years agoHelpfull: Yes(0) No(6)