(x+y+z)5-x5-y5-z5做因式分解

当X=-Y时，原式值为0,因此原式必有因式(X+Y)。原式是对称式，同理也有因式(X+Z),(Y+Z).原式是五次式，而
(X+Y)(Y+Z)(X+Z)是三次式，两者必然相差一个二次对称式：
[K(X^2+Y^2+Z^2)+M(XY+YZ+XZ)]
其中K,M为待定系数。即原式=
(X+Y)(X+Z)(Z+Y)[K(X^2+Y^2+Z^2)+M(XY+YZ+XZ)]
比较等号两边X^4*Y项的系数，知K=5
比较等号两边X^3*Y^2项系数，知K+M=10,M=5,因此原式=

(X+Y)(X+Z)(Z+Y)[5(X^2+Y^2+Z^2)+5(XY+YZ+XZ)]
=5(X+Y)(X+Z)(Z+Y)[(X^2+Y^2+Z^2)+(XY+YZ+XZ)]

=5(X+Y)(X+Z)(Z+Y)(X^2+Y^2+Z^2+XY+YZ+XZ)
分解因式不能带中括号

(x+y+z)5-x5-y5-x5
=(x+y+z)*...(x+y+z)-(x5+y5+z5)
=5(X+Y)(X+Z)(Z+Y)[(X^2+Y^2+Z^2)+(XY+YZ+XZ)]

=5(X+Y)(X+Z)(Z+Y)[(X^2+Y^2+Z^2)+(XY+YZ+XZ)]

(x+y+z)5-x5-y5-z5
=(x+y+z)5-(x5+y5+z5)
=(x+y+z)5-(x+y+z)5
=0