一个因式分解题目轮换法做因式分解(x+y+z)^5-(y+z-x)^5-(z+x-y)^5-(x+y-z)^5

1个回答

  • 令x=0,则

    (x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5

    =(y+z)^5-(y-z)^5-(-y+z)^5-(y+z)^5

    =0

    所以分解式中含因式x,同理可得也含y、z

    从而分解式中含因式xyz

    又因为(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5是五次轮换对称式,

    所以除去xyz外,还有二次轮换对称式,

    只能是k(x^2+y^2+z^2)

    或者k[(x+y)^2+(y+z)^2+(z+x)^2]

    或者k[(x-y)^2+(y-z)^2+(z-x)^2

    用代定系数法可算出k=80,并排除后两种情况,

    可得,原式=80xyz(x^2+y^2+z^2)