# Old problem!Watch

This discussion is closed.
#1
If x1, x2, x3 are the roots of the equation x^3 - x - 1= 0.Calculate
S = (1-x1/1+x1) + (1-x2)/(1+x2) + (1-x3)/(1+x3)
0
14 years ago
#2
(Original post by BCHL85)
If x1, x2, x3 are the roots of the equation x^3 - x - 1= 0.Calculate
S = (1-x1/1+x1) + (1-x2)/(1+x2) + (1-x3)/(1+x3)
Infinite?
0
#3
(Original post by Galois)
Infinite?
I don't think so.
0
#4
yah, I think it should be a number
0
14 years ago
#5
(Original post by BCHL85)
If x1, x2, x3 are the roots of the equation x^3 - x - 1= 0.Calculate
S = (1-x1/1+x1) + (1-x2)/(1+x2) + (1-x3)/(1+x3)
let roots be a,b,c
s= {(1-a)(1+b)(1+c)+(1-b)(1+a)(1+c)+(1-c)(1+a)(1+b)}/(1+a)(1+b)(1+c)
numerator is
(1+c){(1-a)(1+b)+(1-b)(1+a)}+(1-c)(1+a)(1+b)
(1+c){1-a+b-ab+1-b-ab+a}+(1-c)(1+a+b+ab)
(1+c){2-2ab}+(1-c)(1+a+b+ab)
2-2ab+2c-2abc+1+a+b+ab-c-ac-bc-abc
3-ab-ac-bc-3abc+a+b+c (*)

a,b,c roots of x^3-x-1=0
so a+b+c=0
ab+ac+bc=-1
abc=1
(*) becomes 3-(-1)-3=1.
denominator is
(1+a)(1+b)(1+c)=(1+a+b+ab)(1+c)= 1+a+b+ab+c+ac+bc+abc
= 1+a+b+c+ab+bc+ac+abc
=1-1+1
=1
so S=1/1=1.
if there is no error in this i will be amazed!
0
14 years ago
#6
(Original post by BCHL85)
If x1, x2, x3 are the roots of the equation x^3 - x - 1= 0.Calculate
S = (1-x1/1+x1) + (1-x2)/(1+x2) + (1-x3)/(1+x3)
Denote x[a] as x subscript a
________________________________ __________

If x[1], x[2], and x[3] are the roots of the equation x^3 - x - 1 = 0, then

(x - x[1]).(x - x[2]).(x - x[3]) = x^3 - x - 1
=> x^3 - x^2.x[3] - x^2.x[2] + x.x[2].x[3] - x^2.x[1] + x.x[1].x[3] + x.x[2].x[1] - x[1].x[2].x[3] = x^3 - x - 1
=> x^3 - x^2(x[1] + x[2] + x[3]) + x(x[1].x[2] + x[1].x[3] + x[2].x[3]) - x[1].x[2].x[3] = x^3 - x - 1

=> x[1] + x[2] + x[3] = 0
=> x[1].x[2] + x[1].x[3] + x[2].x[3] = -1
=> x[1].x[2].x[3] = 1
________________________________ __________

Now we have

(1-x1/1+x1) + (1-x2)/(1+x2) + (1-x3)/(1+x3)

On expanding this out we get:

(3 + x[3] + x[2] - x[2].x[3] + x[1] - x[1].x[3] - x[1].x[2] - 3.x[1].x[2].x[3])/((1+x[1]).(1+x[2]).(1+x[3]))

=> (3 + 1 - 3)/((1+x[1]).(1+x[2]).(1+x[3]))

=> 1 / (1 + x[3] + x[2]+ x[2].x[3] + x[1] + x[1].x[3] + x[1].x[2] + x[1].x[2].x[3] )

=> 1 / (1 + -1 + 1 )

=> 1 /1

=> 1

Galois.
0
#7
Ok, both of u got it
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