Latest Elitmus Aptitude Question SOLUTION: 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 determine

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(#M40158265) ELITMUS Numerical Ability question Quadratic Equations Keep an EYE

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

Asked In Elitmus Shashank Saha (7 years ago)

Unsolved Read Solution (3)

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saurabh kumar  (7 years ago)

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

Is this solution Helpfull? Yes (8) | No   

Manisha Choudhary  (7 years ago)

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

Is this solution Helpfull? Yes (7) | No   

kamakshi gupta  (7 years ago)

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

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