If 1/a+1/b+1/c= 1/(a+b+c) where a+b+c is not equal to zero,what is the value of (a+b)(b+c)(C+a)
A)equals zero
B) Greater than zero
C) less than zero
D) can not be determined

• i think ....Answer is A
  let a=1,b=-1,c=1
  then the above equation 1/a + 1/b+ 1/c=1/(a+b+c) satisfies
  then (a+b)(b+c)(c+a)=0
  
• 11 years agoHelpfull: Yes(8) No(3)
• ans=(a) equals zero
  =>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=(b^2*c+c^2*b)+(a^2*c+c^2*a)+(a^2*b+b^2*a)+2abc
  = (a+b)(b+c)(c+a)//
  => (a+b)(b+c)(c+a)= 0
  => equal to zero
• 10 years agoHelpfull: Yes(4) No(0)
• @diksha u can call on 9916020682,will explain on phone.
• 11 years agoHelpfull: Yes(3) No(0)
• (b)
  1/a+1/b+1/c= 1/(a+b+c)
  => (a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1
  => [(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=4
  => [(b+c)/a+(a+c)/b+(a+b)/c]+2=4+2
  => [(a+b)(b+c)(C+a)]/abc=6
  => (a+b)(b+c)(C+a)]=6abc
  which is always greater than zero
• 11 years agoHelpfull: Yes(1) No(7)
• put a=-b
  then try
• 10 years agoHelpfull: Yes(1) No(0)
• equal to 0
  
• 10 years agoHelpfull: Yes(1) No(0)
• couldn't understand from step 5 to 6
• 11 years agoHelpfull: Yes(0) No(0)
• will be less than 0
• 11 years agoHelpfull: Yes(0) No(0)
• A)equals zero
  1/a + 1/b + 1/c= 1/(a+b+c)
  divide both side by 'abc'. You will get,
  (bc+ca+ab)(a+b+c) = abc
  multiply the lef side braced elements... you will get...
  a^2c + ac^2 + a^2b + ab^2 + b^2c + bc^2 + 2abc = 0 ------ > 1
  now, multiply the braced elements for which value is being asked... you will get
  the left side elements of the eqution "1" that will give you a value= 0.
• 10 years agoHelpfull: Yes(0) No(0)
• expand (a+b)(b+C)(c+a)
  we get
  2abc+c^2(a+b)+b^2(a+c)+a^2(b+c)
  now solve the 1/a+1/b+1/c= 1/(a+b+c)
  you will get
  2abc=--c^2(a+b)-b^2(a+c)-a^2(b+c)
  on putting this value we get ans=0
• 7 years agoHelpfull: Yes(0) No(0)