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(#M40089746) ELITMUS Numerical Ability question quant Keep an EYE

If 1/a+1/b+1/c=1/(a+b+c) where a+b+c not equal to zero and abc not equal to zero.what is the value of (a+b)(b+c)(c+a)

1)=0 2)>0 3)<0 4) cannot b determined

Asked In Elitmus Vinay (9 years ago)

Unsolved Read Solution (6)

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dipeshshaw  (9 years ago)

Answer is 1)=0
Expalanation:-
1/a+1/b+1/c=1/(a+b+c)
1/a+1/b=1/(a+b+c)-1/c
(b+a)/ab={c-(a+b+c)}/(a+b+c)*c .....................(a+b+c)& abc are non-zero
(a+b)(a+b+c)c = ab{c-(a+b+c)
(a+b)(a+b+c)c=-ab(a+b)
(a+b)(a+b+c)c + ab(a+b)=0
(a+b){(a+b+c)c+ab}=0
(a+b){ac+bc+c2+ab}=0
(a+b){a(c+b)+c(a+b)}=0
(a+b)(b+c)(a+c)=0

Is this solution Helpfull? Yes (32) | No (1)  

vivek  (9 years ago)

1/a+1/b+1/c=1/(a+b+c)
or, (ab+bc+ca)/abc= 1/(a+b+c)
or, (a+b+c)*(ab+bc+ca)=abc
or, a^2b+a^2c+ab^2+ac^2+b^2c+bc^2+3abc=abc
or, a^2b+a^2c+ab^2+ac^2+b^2c+bc^2+2abc=0
or, a^2b+a^2c+ab^2+ac^2+b^2c+bc^2+abc+abc=0
or, (a+b)(b+c)(c+a)=0

Is this solution Helpfull? Yes (3) | No   

kalpana  (9 years ago)

Ans:--- 1)0

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kaustubh salunke  (9 years ago)

0 will be the value.

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akshay  (9 years ago)

0 is write answer by hit n trial method

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Sounak Bhattacharya  (9 years ago)

take a=1 b=1 c=-1 ;)

Answer 0

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