Question

若x1和x2分别是一元二次方程x^2-3x-1=0的两根,求x1^3=x2^3的值

Answer 1

是x1^3+x2^3吧？
x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)=(x1+x2)[(x1+x2)^2-3x1x2]=(x1+x2)^3-3x1x2(x1+x2)
x1^2-3x1-1=0————a
x2^2-3x2-1=0————b
ab两式相减：x1^2-x2^2-3x1+3x2=0
(x1+x2)(x1-x2)-3(x1-x2)=0
(x1-x2)(x1+x2-3)=0
x1=x2舍去
x1+x2-3=0
x1+x2=3
ab相乘
(x1^2-3x1)(x2^2-3x2)=1
(x1x2)^2-3x1x2(x1+x2)+9x1x2=1
把x1+x2=3代入得x1x2=1或-1
若x1x2=1,则x1=1/x2
1/x2+x2=3
x2^2-3x2+1=0与原式不符，故舍去
同理验证得x1x2=-1
代入x1^3+x2^3=(x1+x2)^3-3x1x2(x1+x2)
=3^3-3*-1*3
=27+9
=36

Answer 2

题目没看懂，到底是求什么值啊！