Question

因式分解x^3(y-z)+y^3(z-x)+z^3(x-y)

Answer 1

x^3(y-z)+y^3(z-x)+z^3(x-y)
=x^3(y-z)+y^3z-xy^3+xz^3-yz^3
=x^3(y-z)+(y^3z-yz^3)-(xy^3-xz^3)
=x^3(y-z)+yz(y^2-z^2)-x(y^3-z^3)
=x^3(y-z)+yz(y+z)(y-z)-x(y-z)(y^2+yz+z^2)
=(y-z)[x^3+y^2z-yz^2-x(y^2+yz+z^2)]
=(y-z)(x^3+y^2z-yz^2-xy^2-xyz-xz^2).

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追问

你的答案怎么跟上面完全不一样？

追答

x^3(y-z)+y^3(z-x)+z^3(x-y)
=x^3y-x^3z+y^3z-xy^3+z^3(x-y)
=(x^3y-xy^3)-(x^3z-y^3z)+z^3(x-y)
=xy(x^2-y^2)-z(x^3-y^3)+z^3(x-y)
=xy(x+y)(x-y)-z(x-y)(x^2+xy+y^2)+z^3(x-y)
=(x-y)[xy(x+y)-z(x^2+xy+y^2)+z^3]
=(x-y)(x^2y+xy^2-x^2z-xyz-y^2z+z^3)
=(x-y)[(x^2y-x^2z)+(xy^2-xyz)-(y^2z-z^3)]
=(x-y)[x^2(y-z)+xy(y-z)-z(y^2-z^2)]
=(x-y)[(y-z)(x^2+xy-yz-z^2)]
=(x-y)(y-z)[(x^2-z^2)+(xy-yz)]
=(x-y)(y-z)(x-z)(x+y+z).

x^3(y-z)+y^3(z-x)+z^3(x-y)
=x^3y-x^3z+y^3z-xy^3+z^3(x-y)
=(x^3y-xy^3)-(x^3z-y^3z)+z^3(x-y)
=xy(x^2-y^2)-z(x^3-y^3)+z^3(x-y)
=xy(x+y)(x-y)-z(x-y)(x^2+xy+y^2)+z^3(x-y)
=(x-y)[xy(x+y)-z(x^2+xy+y^2)+z^3]
=(x-y)(x^2y+xy^2-x^2z-xyz-y^2z+z^3)
=(x-y)[(x^2y-x^2z)+(xy^2-xyz)-(y^2z-z^3)]
=(x-y)[x^2(y-z)+xy(y-z)-z(y^2-z^2)]
=(x-y)[(y-z)(x^2+xy-yz-z^2)]
=(x-y)(y-z)[(x^2-z^2)+(xy-yz)]
=(x-y)(y-z)(x-z)(x+y+z).

Answer 2

这是一个关于x、y、z的四次齐次轮换对称式，
当x=y时，原式的值为零，
根据余式定理知x-y是它的一个因式
由轮换对称式的特点
y-z,z-x也是它的因式
3个式子乘起来才3次，还差1次，则必为(x+y+z)
所以原式可以化为k(x+y+z)(x-y)(y-z)(z-x)
取x=2,y=1,z=0，得k=-1
原式=-(x+y+z)(x-y)(y-z)(z-x)

追问

用因式分解的方法，不要用对称式，我想知道用因式分解的方法如何做

追答

(x^3(y-z)+y^3(z-x)+z^3(x-y))
=[xy(x^2-y^2)-z(x^3-y^3)+z^3(x-y)]
=(x-y)(xy(x+y)-z(x^2+y^2+xy)+z^3)
=(x-y)[(x^2y-x^2z)+(xy^2-xyz)-z(y^2-z^2)]
=(x-y)(y-z)(x^2+xy-z(y+z))
=(x-y)(y-z)[(x^2-z^2)+y(x-z)]
=(x-y)(y-z)(x-z)(x+z+y)

追问

上面一步是怎么到下面一步的？

追答

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