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
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