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Algebra

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13 Jan 2007 15:30:47 IST

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COMPLEX NOS

None

IF A, B, C ARE THE CUBE ROOTS OF P (P<0) , THEN FOR ANY x, y, z the expression (xA + yB + zC)/(xB + yC + zA) IS EQUAL TO

a) 1 b) w c) w^2 d) None of these

1, w, w^2 are the cube roots of unity

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Comments (3)

Krishnan Ramachandran

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Joined: 13 Dec 2006

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13 Jan 2007 16:13:30 IST

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The solution to this prob seems to be very easy

Cube root of any number P = P^1/3 . 1^1/3

WHICH MEANS ,

A=P^1/3

B=P^1/3 w

C=P^1/3 w²

therefore in the expresion given , put values of A,B,C and P^1/3 is cancelled out from num and denom.

Take w common from denom . and u get w(num.) as denom.

Answer = 1/w = w²

Das Mukerjee

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13 Jan 2007 23:12:52 IST

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I have a doubt regarding this problem. Since, the expression will have the same value for all x,y,z , let us put arbitarily x=y=z=k, then the value of the expression = k(A+B+C)/k(B+C+A) = 1

Plz tell me why the difference of answers

edison

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14 Jan 2007 12:39:29 IST

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Well done krish. The approach and solution suggested by krish is perfect. And

the answer is undoubtedly w²

Now lets take the QUERY of ariyam66

If we take x = y = z = k (say)

then the expression becomes

(xA + yB + zC)/(xB + yC + zA) = (x + yw + zw²)/(xw + yw² + z)

= k(1 + w + w²) / k (w + w² + 1)

= (1 + w + w²) / (w + w² + 1)

= 0/0 and this can not be equated to 1

As 0/0 is not defined.

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