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Algebra - 1

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Algebra - 1

by Aman verma » Wed Jun 02, 2010 9:29 am
Q: The value of 1/(x^b + x^-c + 1) +1/( x^c + x^-a + 1) + 1/(x^a + x^-b + 1) , given that a + b + c = 0 , is :

a)1

b)0

c)abc

d)x

e)xyz
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by akdayal » Wed Jun 02, 2010 11:57 am
The value of 1/(x^b + x^-c + 1) +1/( x^c + x^-a + 1) + 1/(x^a + x^-b + 1) , given that a + b + c = 0 , is :
1/(x^b + x^-c + 1) = x^c/(x^(b + c) + 1 + x^c) = x^c/(x^-a + 1 + x^c) ; because [b + c = -a]
1/(x^a + x^-b + 1) = x^(b + c)/(x^(a + b+ c) + x ^c + x^(b + c)) = x^-a/(1 + x^c + x^-a)
Add all three term
denominator is same
( 1 + x^c + x^-a)/(1 + x^c + x^-a) = 1
Hence Ans.
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by Aman verma » Thu Jun 03, 2010 2:02 am
OAA.
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by odod » Thu Jun 03, 2010 8:49 am
um can anyone out there explain the above post. I'm having troubles understanding how

1/(x^b + x^-c + 1) = x^c/(x^(b + c) + 1 + x^c) = x^c/(x^-a + 1 + x^c) ;
ODOD
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by odod » Thu Jun 03, 2010 8:51 am
um can anyone out there explain the above post. I'm having troubles understanding how

1/(x^b + x^-c + 1) = x^c/(x^(b + c) + 1 + x^c) = x^c/(x^-a + 1 + x^c) ;
ODOD
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by akdayal » Thu Jun 03, 2010 9:11 am
um can anyone out there explain the above post. I'm having troubles understanding how

1/(x^b + x^-c + 1) = x^c/(x^(b + c) + 1 + x^c) = x^c/(x^-a + 1 + x^c) ;
1/(x^b + x^-c + 1) = 1/(x^b + (1/x^c )+ 1) = 1//((x^b)(x^c) + 1 + x^c)/x^c = x^c/(x^(b + c) + 1 + x^c)
= x^c/(x^-a + 1 + c^c) [b + c = -a given in question]

Hope you will get now
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by RJ Learnz » Thu Jun 03, 2010 9:33 am
As a + b + c = 0 and there is no further constraint Imposed on them,

Just assume a = b = c = 0 for simplicity and substitute values.

Original Eq with this substituition would be:

1/3 + 1/3 + 1/3 = 1

Ans: A
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