已知x-y≠0 x2-x=7 y2-y=7 求x3+y3+x2y+xy2的值

已知x-y≠0 x2-x=7 y2-y=7 求x3+y3+x2y+xy2的值

x²-x=7 y²-y=7
相减
x²-x-y²+y=0
(x+y)(x-y)=x-y
x-y≠0
约分
x+y=1

x²-x=7 y²-y=7
相加
x²+y²-(x+y)=14
(x+y)²-2xy-(x+y)=14
x+y=1
所以xy=-7
x²+y²=(x+y)²-2xy=15

原式=(x³+x²y)+(xy²+y³)
=x²(x+y)+y²(x+y)
=(x²+y²)(x+y)
=15×1
=15

因为x^2-x=7 (1) y^2-y=7 (2)
得x^2-x=y^2-y
整理得： x^2-y^2=x-y
(x+y)(x-y)=x-y
因为x-y不等于0 所以 x+y=1
(1)式+（2）式 得 x^2-x+y^2-y=14
(x^2+y^2)-(x+y)=14
(x+y)^2-2xy-(x+y)=14
因为x+y=1 得 xy=-7
x^3+y^3+x^2y+xy^2
整理得：x^2(x+y)+y^2(x+y)
=(x^2+y^2)(x+y)
=[(x+y)^2-2xy](x+y)
=[1-(-14)]x1
=15

x2-x=7 y2-y=7
x^2-x-(y^2-y)=0
(x-y)(x+y-1)=0
因为：x-y≠0,所以，x+y=1
(x^2-x)+(y^2-y)=(x^2+y^2)-(x+y)=(x^2+y^2)-1
所以，x^2+y^2-1=14
x^2+y^2=15
x3+y3+x2y+xy2
=(x+y)(x^2-xy+y^2)+xy(x+y)
=(x^2-xy+y^2)+xy
=x^2+y^2
=15