因式分解(1+x2)(1+y)^2-(1+y^2)(1+x)2 怎么解？急急急！

设A=(1+y)^2
B=(1+x)^2
原式=（B-2x)A-(A-2y)B=2(yB-xA)然后A，B带入，展开得到
=2（yxx-xyy-x+y)=2(xy(x-y)-(x-y))=2(x-y)(xy-1)

解：
(1+x^2)(1+y)^2-(1+y^2)(1+x)^2
=[(1+x^2)(y^2+2y+1)]-[(1+y^2)(x^2+2x+1)]
=[(y+1)^2+x^2y^2+2x^2y+x^2]-[(x+1)^2+x^2y^2+2xy^2+y^2]
=(y+1)^2-(x+1)^2+2xy(x-y)+x^2-y^2
=(y+1+x+1)(y-x)+2xy(x-y)+(x+y)(x-y)
=(x-y)(-2-x-y+2xy+x+y)
=2(x-y)(xy-1)

题应该是这样吧？~~
(1+x^2)(1+y)^2-(1+y^2)(1+x)^2
=（1+y)^2+（1+y)^2x^2-（1+x)^2-y^2（1+x)^2
=[（1+y)^2-（1+x)^2]+[(1+y)^2x^2-y^2（1+x)^2]
=(1+y+1+x)(1+y-1-x)+(x+xy+y+xy)(x+xy-y-xy)
=(2+y+x)(y-x)+(y+x+2xy)(x-y)
=(y-x)(2+y+x-y-x-2xy)
=2(y-x)(1-xy)

(1+x^2)(1+2y+y^2)-(1+y^2)(1+2x+x^2)
=1+2y+y^2+x^2+2x^2y+x^2y^2-(1+2x+x^2+y^2+2xy^2+x^2y^2）
=-2x+2y+2x^2y-2xy^2
=-2(x-y)+2xy(x-y)
=2(x-y)(xy-1)