据题意得
x^2-y^2=z^2
(x+y)(x-y)=z^2
x^2=y^2+z^2
y^2=x^2-z^2
原式=（x-y）（x^2+xy+y^2)-(x-y)(x+y)z
=(x-y)[x^2+xy+y^2-z(x+y)]
=（x-y)(x^2+xy+y^2-xz-zy)
=(x-y)(x^2+2xy+y^2-xz-xy-zy)
=(x-y)[(x+y)^2-z(x+y)-xy]
.
=(x-z)(x-y)(2x-z+y)
如果x的平方减y的平方减z的平方为0,请把x的立方减y的立方减的立方分解因式