Latest Elitmus Aptitude Question SOLUTION: 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) C

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(#M40155979) ELITMUS Numerical Ability question Keep an EYE

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

Asked In Elitmus insha singh (8 years ago)

Solved Rishu Read Solution (4)

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Read Solution (4) :

Rishu  (8 years ago) Best Solution

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

Is this solution Helpfull? Yes (26) | No (3)  

insha singh  (8 years ago)

its (a+b+c is) not equal to 0 and (abc) is not equal to 0 in the question

Is this solution Helpfull? Yes (3) | No (1)  

Shashank Singh  (7 years ago)

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

Is this solution Helpfull? Yes (2) | No   

manu  (8 years ago)

equal to zero

Is this solution Helpfull? Yes | No (6)  

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