二次根式:已知x=√3-2,求代数式x^4+4x^3-3x²-4x-4的值

x^4+4x^3-4x-4
=x^4+2x^3+2x^3-4x-4
=x³(x+2)+2x^3-4x-4
=x³[(x+2)+2]-4x-4
=x³[(√3-2+2)+2]-4x-4
=x²*x(√3+2)-4x-4
=x²(√3-2)(√3+2)-4x-4
=x²(3-4)-4x-4
=-x²-4x-4
=-(x+2)²
=-(√3-2+2)²
=-3问题是求代数式x^4+4x^3-3x2-4x-4的值，你看错题了x^4+4x^3-3x²-4x-4=x^4+2x^3+2x^3-2x²-x²-4x-4=x^3(x+2)+2x^3-2x²-(x+2)²=x^3(√3-2+2)+2x^3-2x²-(√3-2+2)²=x³(√3+2)-2x²-3=x²*x*(√3+2)-2x²-3=x²*(√3-2)*(√3+2)-2x²-3=-x²-2x²-3=-3x²-3=-3(√3-2)²-3=-3(3+4-4√3)-3=-24+12√3.