（1+y）^2-2x^2(1+y^2)+x^4(1-y)^2

原式=[(1+y)-x^2(1-y)]^2-4*x^2*y^2
=(1+y-x^2+x^2*y)^2-(2xy)^2
=(1+y+2xy-x^2+x^2*y)(1+y-2xy-x^2+x^2*y)
=[(1-x^2)+(yx+y)+(x^2*y+xy)][(1-x^2)-(xy-y)+(x^2*y-xy)]
=(x+1)(x-1)(x-xy+1+y)(x-xy-1-y)
这道体表面上是不可以分解的，因为这里是1+y^2，而不是1-y^2，所以不能用完全平方公式，那么，就暂且把题目看作是（1+y）^2-2x^2(1-y^2)+x^4(1-y)^2，改了以后的题目要和原题相等就还要减去4*x^2*y^2，4*x^2*y^2就等于(2xy)^2，接下来就可以用平方差公式了。

（1+y）^2-2x^2(1+y^2)+x^4(1-y)^2=
=1+2y+y^2-2x^2-2x^2y^2+x^4-2x^4y+x^4y^2=
=2y(1-x^4)+(1-x)-x(1-x^3)+y^2(1-x^2)=
=2y(1-x^4)+(1-x)-x(1-x^3)+y^2(1-x^2)=
=2y(1-x)(1+x)(1+x^2)+(1-x)-x(1-x)(1+x+x^2)+y^2(1-x)(1+x)=
=(1-x)[2y(1+x)(1+x^2)+1-x(1+x+x^2)+y^2(1+x)].
完成因式分解！

(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2
=(1+y)^2-2x^2(1+y)(1-y)+x^4(1-y)^2-4x^2y^2
=[(1+y)-x^2(1-y)]^2-4x^2y^2
=(1+y-x^2+x^2y)^2-(2xy)^2
=(1+y+2xy-x^2+x^2y)(1+y-2xy-x^2+x^2y)
=[(1-x^2)+(yx+y)+(x^2y+xy)][(1-x^2)-(xy-y)+(x^2y-xy)]
=(x+1)(x-1)(x-xy+1+y)(x-xy-1-y)