Question

50分 急！！求解一道初中数学题

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

据说是轮换对称式，

步骤能详细的详细点
分解因式

写出步骤

Answer 1

由题 式可以 看出 当x=y 或 y=z 或x=z时式子为0 所以肯定有因式(x-y)(y-z)(z-x) 展开后x最高项为-x^2 y 与 x^2z 而原式中x最高次项为x^3y和-x^3z 所以还差x的1次项因式，所以实际缺的是-(x+y+z)分解因式为
-（x-y)(y-z)(z-x)(x+y+z)

Answer 2

x^3(y-z)+y^3(z-x)+z^3(x-y)

Answer 3

x换成y，y换成z，z换成x
yz(y^2-z^2)+zx(z^2-x^2)x+y(x^2-y^2)

多项式不变

Answer 4

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)
可以仔细观察最后化简完的式子，即x^3(y-z)+y^3(z-x)+z^3(x-y)，共三项，每一项都有x y z 三个字母，且轮换着3次方，其余两个字母就做减法，这就是轮换对称式 希望可以帮助到你哦

Answer 5

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

Answer 6

确实是轮换对称式,
将x=y代入时,为0，
将x=z代入时,为0，
将z=y代入时,为0，
说明有因式x-y,y-z,z-x,
由于是四次多项式,还有因式X+Y+Z,
所以(x+y+z)(x-y)(y-z)(z-x)