Given x_1,x_2,x_3,x_4 the roots of p(x) = x^4+x^3+x^2+x+1 = 0 calculate x_1^8+x_2^(18)+x_3^(28)+x_4^(38) ?
2 Answers
Explanation:
Given:
p(x) = x^4+x^3+x^2+x+1
Note that:
(x-1)p(x) = (x-1)(x^4+x^3+x^2+x+1) = x^5-1
So we can tell that the zeros of
So if the zeros of
x_1^5=x_2^5=x_3^5=x_4^5=1
So:
x_1^8+x_2^18+x_3^28+x_4^38 = x_1^3+x_2^3+x_3^3+x_4^3
Now we have a symmetric polynomial in the zeros, we can express it in terms of the elementary symmetric polynomials that form the coefficients of
Note:
p(x) = (x-x_1)(x-x_2)(x-x_3)(x-x_4)
color(white)(p(x)) = x^4-(x_1+x_2+x_3+x_4)x^3+(x_1x_2+x_2x_3+x_3x_4+x_4x_1)x^2-(x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2)x+x_1x_2x_3x_4
Equating coefficients, we have:
{ (x_1+x_2+x_3+x_4 = -1), (x_1x_2+x_2x_3+x_3x_4+x_4x_1=1), (x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2=-1), (x_1x_2x_3x_4=1) :}
Now:
x_1^2+x_2^2+x_3^2+x_4^2
=(x_1+x_2+x_3+x_4)^2-2(x_1x_2+x_2x_3+x_3x+x_4x_1+x_1x_3+x_2x_4)
=(-1)^2-2(1)
=-1
Note that:
-1 = (x_1+x_2+x_3+x_4)(x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4)
color(white)(-1) = (x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4)+3(x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2)
color(white)(-1) = (x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4)+3(-1)
Hence:
x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4 = 2
Then:
1 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2)
color(white)(1)=(x_1^3+x_2^3+x_3^3+x_4^3)+(x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4)
color(white)(1)=(x_1^3+x_2^3+x_3^3+x_4^3)+2
So:
x_1^3+x_2^3+x_3^3+x_4^3 = -1
See below.
Explanation:
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