分解因式:(y-z)^5+(z-x)^5+(x-y)^5

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)^5+(x-y)^5=5（x-y）（y-z）（z-x）[x^2+y^2+z^2-xy-yz-xz]

(y-z)^5+(z-x)^5+(x-y)^5
=a^5+b^5+c^5
=a^5+b^5-(a+b)^5
=a^5+b^5-(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5)
=-5(a^4b+ab^4+2a^3b^2+2a^2b^3)
=-5ab(a^3+b^3+2a^2b+2ab^2)
=-5ab(a^2(a+b)+b^2(a+b)+a^2b+ab^2)
=-5ab(a+b)(a^2+b^2+ab)
=5(x-y)(y-z)(z-x)((x-y)^2+(y-z)^2+(x-y)(y-z))
=5(x-y)(y-z)(z-x)(x^2+y^2+z^2-xy-yz-zx)