x^2(y+z)+y^2(z+x)+z^2(x+y)+2xy

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

我據
答案( X+Y ) ( Y+Z) ( Z+X )
譯為 = = + = = + = =
粵語(巡壹) (宇宙) (之凶)

另一方法

Let f(x,y,z) = x²(y+z)+y²(z+x)+z²(x+y)+2xyz

f(x,y,0) = x²(y+0)+y²(0+x)+0²(x+y)+2xy0
= x²y+y²x

f(x,-x,0) = x²(-x)+(-x)²x = 0

Therefore, (x+y) is a factor of f(x,y,z).

2013-08-29 15:34:46 補充：
Note that f(x,y,z) is cyclic symmetric (i.e., f(x,y,z) = f(y,z,x)), therefore, (y+z) and (z+x) are also factor of f(x,y,z).

Consider f(x,y,z) = k(x+y)(y+z)(z+x).

By comparing coefficient, k = 1, and the answer is the same as that by 螞蟻雄兵.