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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