2(x+y)(x+y-z)+3(y-z)(x+y-z)

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2(x+y)(x+y-z)+3(y-z)(x+y-z)
2(x+y)(x+y-z)+3(y-z)(x+y-z)

2(x+y)(x+y-z)+3(y-z)(x+y-z)
(x+y-z)(2x+5y-3z)
是不是这个答案

因为原式为x,y,z的5次轮换式当x=y时原式为0
（x-y）（y-z）（z-x）为因式设:(y-z)^5+(z-x)^5+(x-y)^5=（x-y）（y-z）（z-x）[l（x^2+y^2+z^2）+m(xy+yz+xz)]
令x=1 y=2 z=0 5l+2m=15
令x=1 z=-1 y=0 2l-m=15
l=5 m=-5
(y-z)^5+(z-x)...

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