当m为何值时 x³+y³+z³+mxyz能被x+y+z整除

当m为何值时 x³+y³+z³+mxyz能被x+y+z整除
当m为何值时 x³+y³+z³+mxyz能被x+y+z整除

当m为何值时 x³+y³+z³+mxyz能被x+y+z整除
可以这样想,就是设法提出x+y+z出来,采用降次的方法,
x＾3+y＾3+z＾3
=x＾2(x+y+z)+y＾2(x+y+z)+z＾2(x+y+z)-x＾2(y+z)-y＾2(x+z)-z＾2(x+y)
=(x＾2+y＾2+z＾2)(x+y+z)-x＾2(y+z)-y＾2(x+z)-z＾2(x+y)
=(x＾2+y＾2+z＾2)(x+y+z)-[xy(x+y+z)+xz（x+y+z)+yz(x+y+z)-3xyz]
=(x＾2+y＾2+z＾2)(x+y+z)-（xy+xz+yz)(x+y+z）+3xyz
故当m=-3时,代数式能被x+y+z整除