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Cubic Equation Roots

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By Maria Jose Leon-Sotelo Esteban on Monday, January 21, 2002 - 07:11 am:

If a,b, and c are the roots of x³-x-1=0, compute:
(1+a)/(1-a) + (1+b)/(1-b) + (1+c)/(1-c).

Thanks.
M.J.

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By David Loeffler on Monday, January 21, 2002 - 09:22 am:

Well, look at it this way:

x³-x-1 = (x-a)(x-b)(x-c)

x³-x-1 = x³ - (a+b+c)x² +(ab+bc+ca)x - abc

So a+b+c = 0
ab+bc+ca = -1
abc = 1

Now, you should be able to show (by brute force) that if you multiply up the given expression both numerator and denominator may be written solely in terms of these three quantities, and thus obtain the answer. (I make it -7.)

David

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