Latest Elitmus Aptitude Question SOLUTION: 1/a+1/b+1/c=1/(a+b+c) find (a+b)(b+c)(c+a)

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1/a+1/b+1/c=1/(a+b+c) find (a+b)(b+c)(c+a)

Asked In Elitmus mayank (10 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

RAKESH  (10 years ago)

1/a+1/b+1/c=1/(a+b+c)
=> (bc+ca+ab)(a+b+c)= abc
=> abc+bc(b+c)+ bca+ca(c+a) +abc+ab(a+b)= abc
=> bc(b+c) + c^2.a + c.a^2 + a^2.b + a.b^2 + 2abc =0
=> bc(b+c) + (c.a^2 + a^2.b) + (c^2.a + a.b^2 + 2abc) =0
=> bc(b+c) + a^2(b+c) + a(b+c)^2 = 0
=> (b+c)[bc+a^2 + ab+ac]=0
=> (b+c)[(bc+ac)+(a^2+ab)]=0
=> (b+c)[c(a+b)+a(a+b)]=0
=> (a+b)(b+c)(c+a)=0

2nd mtd

put a=1,b=-1,c=1 such that a,b,c & (a+b+c) not equals zero & 1/a+1/b+1/c=1/(a+b+c)

so (a+b)(b+c)(c+a)=(1-1)(-1+1)(1+1)=0

Is this solution Helpfull? Yes (31) | No   

MADHU  (10 years ago)

1/a+1/b+1/c = 1/(a+b+c)
=> [(ab+bc+ca)/(abc)]= 1/(a+b+c)
=> a^2.b+a.b^2+b^2.c+b.c^2+a^2.c+a.c^2+3abc=abc
=> (a^2.b+a.b^2+b^2.c+b.c^2+a^2.c+a.c^2+2abc)=0
since [(a^2.b+a.b^2+b^2.c+b.c^2+a^2.c+a.c^2+2abc)= (a+b)(b+c)(c+a)]
=> (a+b)(b+c)(c+a)=0
ans is 0

Is this solution Helpfull? Yes (6) | No   

anuj kumar jayswal  (10 years ago)

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

Is this solution Helpfull? Yes (4) | No   

PRADEEP KUMAR  (10 years ago)

1/a+1/b+1/c=1/(a+b+c)
(bc+ac+ab)/abc=1/(a+b+c)
(a+b+c)(bc+ac+ab)=abc
abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2+abc=abc
abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2=0
(a+b)(b+c)(c+a)=0
so the ans will be(=0)

Is this solution Helpfull? Yes (1) | No   

santosh guruju  (10 years ago)

the only possible solution is
for iit a=1 ,r=2
for iim a=1 ,d=2
so
iit->1+2+4+8=15
iim->1+3+5+7=16
total score till first half is
1+2+1+3=7

Is this solution Helpfull? Yes | No (1)  

Ajoy Singh  (10 years ago)

its 0
coz
1/a+1/b+1/c=1/(a+b+c)
=>abc+a^2c+ba^2+b^2c+abc+b^2a+bc^2+ac^2+abc=abc
=>(b+a)(b+c)(c+a)=0

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