已知：x+y-2=0,求（x方-y方）的平方-8（x方+y方）的值

x+y=2
(x+y) 2=4
x2+y2+2xy=4
(x2+y2)2=(4-2xy)2
即x4+2x2y2+y4=16-16xy+4x2y2 则x4+y4=16-16xy+2x2y2 (1)
(x2-y2)2-8(x2+y2)
=x4-2x2y2+y4-8x2-8y2
将（1）代入
则原式=16-16xy+2x2y2-2x2y2-8x2-8y2
=16-(16xy+8x2+8y2)
=16-8(x+y) 2
=16-8*4
=-16
其实这类题只要注意化简已知条件即可O(∩_∩)O~

由已知得x+y=2
(x^2-y^2)^2-8(x^2+y^2)
=(x-y)^2*(x+y)^2-8(x^2+y^2)
=4(x-y)^2-8(x^2+y^2)
=4(x^2-2xy-y^2-2x^2-2y^2)
=-4(x+y)^2
=-16

负六十四