问大家一道题，已知分式(x^2-x+1)⼀x=3,试求x^2⼀(x^4+x^2+1)的值。谢谢！！！

(x^2-x+1)/x=3化简得x+1/x=4,所以（x+1/x）^2=4^2,展开整理得到x^2+1/x^2=14
x^2/(x^4+x^2+1)分子和分母同时除以x^2得1/（x^2+1+1/x^2）=1/(1+14)=1/15

(x^2-x+1)/x=3
==》x^2-x+1=3x
==>x^2=4x-1

x^2/(x^4+x^2+1)
=(4x-1)/[x^2(x^2+1)+1]
=(4x-1)/[(4x-1)*4x+1]
=(4x-1)/[16x^2-4x+1]
=(4x-1)/[16(4x-1)-4x+1]
=(4x-1)/[60x-15]
=1/15

(x^2-x+1)/x=x+1/x-1=3，x+1/x=4，(x+1/x)^2=x^2+1/x^2+2=16，x^2+1/x^2=14。
x^2+1/x^2+1=(x^4+x^2+1)/x^2=15，x^2/(x^4+x^2+1)=1/15。

x-1+1/x=3
x+1/x=4
原式=1/(x^2+1+1/x^2) ------x^2+1/x^2=(x+1/x)^2-2=4^2-2=14
=1/(14+1)
=1/15