设x1、x2是二次方程x2+x-3=0的两个根,求x1的三次方-4倍的x2的平方+2028的值

x1^3+x1^2-3x1=0,x1^3=3x1-x1^2
x1+x2=-1 (1)
x1*x2=-3 (2)
(1)^2 x1^2+x2^2+2x1x2=1
(2)代入 x1^2+x2^2=7
4x2^2=27-4x1^2
x1^3-4x2^2+2028=x1^3-27+4x1^2+2028
=x1^3+4x1^2+2001
=3x1-x1^2+4x1^2+2001
=3x1^2+3x1+2001
=3(x1^2+x1)+2001
=3*3+2001
=2010