X3-3X+1=0 求X3+1/X3

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X3-3X+1=0 求X3+1/X3

X3-3X+1=0 求X3+1/X3
X3-3X+1=0 求X3+1/X3

X3-3X+1=0 求X3+1/X3
因为x^3-3x+1=0 所以x^3-3x=-1 x^3+1=3x 1/x=3-x^2(两边同除x)
分别用这几个式子的变形进行整体代换
x^3+1/x^3
=3x-1+(3x-x^3)/x^3(因为x^3=3x-1,1=3x-x^3)
=3x-1+3/(x^2)-1
=3(x^3+1)/x^2-2(因为x^3+1=3x)
=3*3x/x^2 -2
=9/x-2.①
=9*(3-x^2)-2=27-9x^2-2=25-9x^2.②
因为x^3-3x+1=0
so (x^2-3x+1)^2=0
x^6-3x^4+x^3-3x^4+9x^2-3x+x^3-3x+1=0
x^6-6x^4+2x^3+9x^2-6x+1=0
x^6-6x^4=-2x^3-9x^2+6x-1.③
x^3+1/x^3
=(3-x^2)^3+x^3
=x^6-3x^4-6x^4+18x^2+9x^2-27+x^3
=x^6-9x^4+27x^2-27+x^3
=x^6-6x^4-3x^4+27x^2-27+x^3 ④
③代入④得
x^3+1/x^3
=-2x^3-9x^2+6x-1-3x^4+27x^2-27+x^3
=-3^4+18x^2+6x-28-x^3
=-3x*(x^3-3x)+9x^2+6x-28-x^3(因为x^3-3x=-1)
=3x+9x^2+6x-28-x^3
=9x^2+9x-28-x^3 .⑤
因为⑤=② 所以9x^2+9x-28-x^3=25-9x^2
整理得到-x^3+18x^2+9x-53=0.⑥
又由题设x^3-3x+1=0.⑦
⑥+⑦得18x^2+6x-54=0
3x^2+x-9=0
解两元一次方程
解得x=(-1+√109)/6or(-1-√109)/6
代入②所以25-9x^2=(-5+√109)/2or(-5-√109)/2
即x^3+1/x^3=(-5+√109)/2or(-5-√109)/2