已知X1,X2 是方程X的平方+3X+1=0的两根,则x1的三次方+8x2的平方+20=

X1、X2是方程X^2+3X+1=0的两实数根
韦达定理得:
X1+X2=-3
X1X2=1
X1^2+3X1+1=0
x1^2=-(3x1+1)
x1^3+8x2+20
=-x1*(3x1+1)+8x2+20
=-3x1^2-x1+8x2+20
=-3*(x1^2)-x1+8x2+20
=3*(3x1+1)-x1+8x2+20
=8(x1+x2)+23
=-24+23
=-1