题目

没有学韦达定理 请用易懂步骤回答
已知x-y≠0,x^2-x=7,y^2-y=7,求x^3+y^3+x^2y+xy^2的值.

提问时间：2021-01-19

答案

x^2-x=7
y^2-y=7
两式相减
得x^2-y^2-x+y=0
(x-y)(x+y)-(x-y)=0
(x+y-1)(x-y)=0
因为x-y≠0
所以x+y=1
x^3+y^3+x^2y+xy^2
=(x+y)(x^2+y^2-xy)+xy(x+y)
=1*(x^2+y^2-xy)+xy*1
=x^2+y^2
[x^2-x=7 x^2=x+7
y^2-y=7 y^2=y+7]
=x+7+y+7
=x+y+14
=1+14
=15