Question

分解因式：yz(y-z)+zx(z-x)+xy(x-y)

Answer 1

f(x,y,z) =yz(y-z)+zx(z-x)+xy(x-y)
f(x,x,z) =xz(x-z)+zx(z-x)+x^2.(x-x) =0 => (x-y) is a factor
f(x,y,y) =xy(y-y)+yx(y-x)+xy(x-y)=0 => (y-z) is a factor
f(z,y,z) =yz(y-z)+z^2.(z-z)+zy(z-y)=0 => (z-x) is a factor
=>

yz(y-z)+zx(z-x)+xy(x-y) = k(x-y)(y-z)(z-x)
coef. of y^2.(z-x) => k=-1
ie
yz(y-z)+zx(z-x)+xy(x-y) = -(x-y)(y-z)(z-x)

Answer 2

原式= y²z - yz² + z²x - zx² + xy(x - y)
= z²(x - y) - z(x² - y²) + xy(x - y)
= z²(x - y) - z(x + y)(x - y) + xy(x - y)
= (x - y)(z² - zx - zy + xy)
= (x - y)(z(z - x) - y(z - x))
= (x - y)(z - x)(z - y)

Answer 3

分解因式：
yz(y-z)+zx(z-x)+xy(x-y)
=y²z-yz²+xz²-x²z+x²y-xy²

Answer 4

xy(x-y)+yz(y-z)+zx(z-x)
=xy(x-y)+yz(y-z)+zx[-(x-y)-(y-z)]
=(x-y)(xy-xz)+(y-z)(yz-zx)
=(x-y)(y-z)x+(y-z)(y-x)z
=(x-y)(y-z)(x-z)