=[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz
=[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)
=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)
=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z)
=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)
分解因式x3+y3+z3-3xyz
解：x^3+y^3+z^3-3xyz =[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz =[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z)=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)
x3+y3+z3-3xyz 和 x3+y3+z3+3xyz分解因式
解：x^3+y^3+z^3-3xyz ==[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz =[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z)=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)用到...
分解因式x3+y3+z3-3xyz
x^3+y^3+z^3-3xyz =[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz =[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z)=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)
x^3+y^3+z^3-3xyz因式分解
x^3+y^3+z^3-3xyz = (x^3+3yx^2+3xy^2+y^3)+z^3-3xyz-3yx^2-3xy^2 = (x+y)^3+z^3-3xy(x+y+z)= (x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)= (x+y+z)[(x^2+2xy+y^2-xz-yz+z^2)-2xy]= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)不懂还可...
x^3+y^3+z^3 -3xyz 这个怎么分解因式 各位老师帮帮忙
= (x+y)[(x+y)^2-3xy]+z^3-3xyz = (x+y)^3-3xy(x+y)+z^3-3xyz = (x+y)^3+z^3-3xy(x+y)-3xyz = (x+y+z)[(x+y)^2+z^2-z(x+y)]-[3xy(x+y)+3xyz]= (x+y+z)(x^2+y^2+2xy+z^2-xz-yz)-3xy(x+y+z)= (x+y+z)(x^2+y^2+z^2-xy-xz...
x^3+y^3+z^3-3xyz(分解因式)
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)由x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2）-z（x^2+y^2）-x（y^2+z^2）-y（x^2+z^2）x^3+y^3+z^3-3xyz=……=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)...
x的立方+y的立方+z的立方-3xyz 因式分解,急~~~
x^3+y^3+z^3-3xyz =[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz =[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z)=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)
x^3+y^3+z^3-3xyz用因式定理的思路因式分解
尝试得(x+(y+z))为原式的因式，因为f(-(y+z))=-(y+z)^3+(3yz)(y+z)+y^3+z^3=0 于是用大除法计算(x^3-(3yz)x+y^3+z^3)\/(x+(y+z))，得到另一因式为x^2-(y+z)x+y^2+z^2-yz 最后整理得到(x+y+z)(x²+y²+z²-xy-xz-yz)...
已知:x+y+z=0,求证:x^3+y^3+z^3=3xyz
因式分解 x^3+y^3+z^3-3xyz=（x+y+z）(x^2+y^2+z^2-xy-yz-zx)=0，所以原式成立
问一些初1数学题,因式分解的
学过因式分解，应该知道公式 x^3+y^3+z^3-3xyz=(y+x+z)*(x^2+y^2+z^2-xy-yz-xz)，因为x+y+z=0，所以x^3+y^3+z^3-3xyz=0，所以xyz=(x^3+y^3+z^3)\/3=0。a\/b-b\/a-(a2+b2)\/ab=a\/b-b\/a-(a\/b+b\/a)=-2b\/a，3a2+ab-2b2=0，两边除以a2（显然，由题意，...
