已知x^2+y^2=z^2,xy=10,求（x+y+z)(x+y-z)(x-y+z)(-x+y+z)

(x+y+z)(x+y-z)(x-y+z)(-x+y+z)
=[(x+y)^2-z^2][z^2-(x-y)^2]
=(x^2+2xy+y^2-z^2][z^2-x^2+2xy-y^2]
=2xy*2xy
=4(xy)^2
=400

x^2+y^2-z^2=0
z^2-x^2-y^2=0

原式=[(x+y)+z][(x+y)-z][z+(x-y)][z-(x-y)]
=[(x+y)^2-z^2][^2z-(x-y)^2]
=(x^2+y^2+2xy-z^2)(z^2-x^2-y^2+2xy)
=(0+2xy)(0+2xy)
=4(xy)^2
=400

（x+y+z)(x+y-z)(x-y+z)(-x+y+z)=（x+y+z)(x+y-z) * -(x-y+z)(x-y-z)
=[(x+y)^2-z^2]*-[(x-y)^2-z^2]
(x+y)^2-z^2=2xy=20
(x-y)^2-z^2=-2xy=-20
原式=20*20=400