Question

分解因式，帮我！！xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)

说明一下方法谢谢

Answer 1

xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)
=x^3y-xy^3+y^3z-yz^3+z^3x-zx^3
=x^3(y-z)+y^3(z-x)+z^3(x-y)
因为x-y=(x-z)+(z-y),所以：
=x^3(y-z)+y^3(z-x)+z^3[(x-z)+(z-y)]
=(x^3-z^3)(y-z)+(y^3-z^3)(z-x)
=(x-z)(x^2+xz+z^2)(y-z)+(y-z)(y^2+yz+z^2)(z-x)
=(x-z)(y-z)(x^2+zx+z^2-y^2-yz-z^2)
=(x-z)(y-z)(x-y)(x+y+z)
另外，观察可知它是轮换式，根据轮换式的做法，
可以设为：=k(x-y)(y-z)(z-x)(x+y+z)
待定系数法~

Answer 2

这是一个轮换对称式,易知,它可分解为k(x-y)(y-z)(z-x)(x+y+z),令x=1,y=2,z=3代入,可知k=-1.

Answer 3

原式 = x^3(y - z)- x(y^3 - z^3) + yz(y^2 - z^2)
= (y - z)(x^3 -x(y^2 + z^2 + xy) + (y + z)yz)
= (y - z)(x(x^2 -z^2) - y^2(x - z) - y(x^2-z^2))
= (y - z)(x - z)(x(x + z) - y^2 - y(x + z))
= (y - z)(x - z)(x^2 + zx + z^2 - y^2 -yz - z^2)
= (x - z)(y - z)(x - y)(x + y + z)