x(y-z)^3+y(z-x)^3+z(x-y)^3
2个回答
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令x=y
则原式=y(y-z)^3+y(z-y)^3+z(y-y)^3
=y(y-z)^3-y(y-z)^3
=0
则x-y是x(y-z)^3+y(z-x)^3+z(x-y)^3的一个因式
同理,y-z,z-x,也是x(y-z)^3+y(z-x)^3+z(x-y)^3的一个因式
由于式子具有轮转对称性,猜剩余一个因式为k(x+y+z)
x(y-z)^3+y(z-x)^3+z(x-y)^3=k(x+y+z)(x-y)(y-z)(z-x)
易得,k=1
则
x(y-z)^3+y(z-x)^3+z(x-y)^3=(x+y+z)(x-y)(y-z)(z-x)
则原式=y(y-z)^3+y(z-y)^3+z(y-y)^3
=y(y-z)^3-y(y-z)^3
=0
则x-y是x(y-z)^3+y(z-x)^3+z(x-y)^3的一个因式
同理,y-z,z-x,也是x(y-z)^3+y(z-x)^3+z(x-y)^3的一个因式
由于式子具有轮转对称性,猜剩余一个因式为k(x+y+z)
x(y-z)^3+y(z-x)^3+z(x-y)^3=k(x+y+z)(x-y)(y-z)(z-x)
易得,k=1
则
x(y-z)^3+y(z-x)^3+z(x-y)^3=(x+y+z)(x-y)(y-z)(z-x)
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嘻嘻,,看不懂我是初一的
追答
你们分解因式没有讲
轮转对称法
么?
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x(y-z)^3+y(z-x)^3+z(x-y)^3
=x(y-z)^3+y(z-y)+(y-x)^3+z(x-y)^3
=x(y-z)^3+y(z-y)^3+3y(z-y)^2(y-x)+3y(z-y)(y-x)^2+y(y-x)^3+z(x-y)^3
=(x-y)(y-z)^3+3y(z-y)(y-x)(z-x)+(z-y)(x-y)^3
=(x-y)(y-z)((y-z)^2+3y(z-x)-(x-y)^2)
=(x-y)(y-z)((y-z)^2-(x-y)^2+3y(z-x)这里有个平方差公式y-z+x-y=x-z正好可以提
=(x-y)(y-z)(z-x)(-(y-z-x+y)+3y)
=(x-y)(y-z)(z-x)(x+y+z)
=x(y-z)^3+y(z-y)+(y-x)^3+z(x-y)^3
=x(y-z)^3+y(z-y)^3+3y(z-y)^2(y-x)+3y(z-y)(y-x)^2+y(y-x)^3+z(x-y)^3
=(x-y)(y-z)^3+3y(z-y)(y-x)(z-x)+(z-y)(x-y)^3
=(x-y)(y-z)((y-z)^2+3y(z-x)-(x-y)^2)
=(x-y)(y-z)((y-z)^2-(x-y)^2+3y(z-x)这里有个平方差公式y-z+x-y=x-z正好可以提
=(x-y)(y-z)(z-x)(-(y-z-x+y)+3y)
=(x-y)(y-z)(z-x)(x+y+z)
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