數學知識交流---展開

(w+x-y)(w-x+y)(x+y-w)(w+x+y)
=(w+x+y)(w+x-y)(w-x+y)(x+y-w)
=-[(w+x)+y][(w+x)-y][(w-x)+y][(w-x)-y]
=-[(w+x)^2-y^2][(w-x)^2-y^2]
=-(w^2-x^2)^2+y^2(w+x)^2+y^2(w-x)^2-y^2
=-w^4+2w^2x^2-x^4+y^2(w^2+2wx+x^2+w^2-2wx+x^2-1)
=-w^4+2w^2x^2-x^4+2w^2y^2+2x^2y^2-y^2

(w+x-y)(w-x+y)(x+y-w)(w+x+y)
=[w+(x-y)][w-(x-y)][(x+y)-w][(x+y)+w]
=[w^2-(x-y)^2][(x+y)^2-w^2]
=-[w^2-(x-y)^2][w^2-(x+y)^2]
=-{w^4-[(x-y)^2+(x+y)^2]w^2+(x-y)^2(x+y)^2}
=-w^4+[(x-y)^2+(x+y)^2]w^2-(x-y)^2(x+y)^2

2011-07-21 17:12:23 補充：
=-w^4+[(x^2-2xy+y^2)+(x^2+2xy+y^2)]w^2-[(x-y)(x+y)]^2
=-w^4+2(x^2+y^2)w^2-(x^2-y^2)^2
=-w^4+2(w^2)(x^2)+2(w^2)(y^2)-[x^4-2(x^2)(y^2)+y^4]
=-w^4+2(w^2)(x^2)+2(w^2)(y^2)-x^4+2(x^2)(y^2)-y^4
=2(w^2)(x^2)+2(x^2)(y^2)+2(y^2)(w^2)-w^4-x^4-y^4

2011-07-21 17:31:14 補充：
w, x and y are cyclic
w->x x->y y->w

w+x-y -> x+y-w -> y+w-x
w+x+y -> w+x+y

2(w^2)(x^2)-w^4 -> 2(x^2)(y^2)-x^4 -> 2(y^2)(w^2)-y^4