Question

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

Answer 1

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