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Let a, b, c be the roots of x^3 - x - 1 = 0.  Compute

(1 + a)^2/(1 - a) + (1 + b)^2/(1 - b) + (1 + c)^2/(1 - c).

 Dec 31, 2020
 #1
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thanks my issue has been fixed.

 Dec 31, 2020
 #2
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About finding general an+bn+cn,an+bn+cn, there is an elementary aspect to this, although not necessarily what you want. You already know that a+b+c=0,a+b+c=0, and the other answer gives enough to find a2+b2+c2a2+b2+c2 and a3+b3+c3.a3+b3+c3. Call those x1,x2,x3,x1,x2,x3, then solve in perpetuity with

xn+3=xn+1+xnxn+3=xn+1+xn

Two errors in the question,

bc+ca+ab=−1,bc+ca+ab=−1,

abc=1.abc=1.

So far, I get

x1=0,x1=0,

x2=2,x2=2,

x3=3,x3=3,

x4=2,x4=2,

x5=5,x5=5,

x6=5,x6=5,

x7=7,x7=7,

x8=10,x8=10,

x9=12,x9=12,

x10=17,x10=17,

Since the real root of x3−x−1x3−x−1 is slightly larger than 1,1, the numbers increase, are positive and so on. The two complex roots are smaller than 11 in modulus, so, for large n,n, we get xn≈Rnxn≈Rn where R≈1.3247R≈1.3247 is the real root. For example, R10≈16.643

For more in debt answer just ask (hopefully its debt enough through)

-wolfie

 Jan 1, 2021

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