二次根式:已知x=√3-2，求代数式x^4+4x^3-3x눀-4x-4的值

x^4+4x^3-3x²-4x-4
=x^3(x+4)-3x²-4x-4
=x^3(√3-2+4)-3x²-4x-4
=x^3(√3+2)-3x²-4x-4
=x^2[x(√3+2)-3]-4x-4
=x^2[(√3-2)(√3+2)-3]-4x-4
=x^2[(3-4)-3]-4x-4
=x^2(-4)-4x-4
=-4(x^2+x+1)
=-4[x(x+1)+1]
=-4[(√3-2)(√3-2+1)+1]
=-4[(√3-2)(√3-1)+1]
=-4[(3-2√3-√3+2)+1]
=-4(3-3√3+3)
=-4(6-3√3)
=-12(2-√3)
=-12(2-1.732)
=-12X0.268
=--3.216

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

计算器已经发明了这么多年了，为什么我们的学生还要进行这种浪费时间精力的作业呢？真无聊