= (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)
不懂还可问，满意请及时采纳！o(∩_∩)o
=x³+3x²y+3xy²+y³+z³-3x²y-3xy²-3xyz
=(x+y)³+z³-3xy(x+y+z)
=(x+y+z)(x²+2xy+y²-xz-yz-3xy)
=(x+y+z)(x²+y²+z²-xy-yz-xz)来自：求助得到的回答本回答被提问者采纳
= (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)
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...
x^3+y^3+z^3-3xyz因式分解
x³+y³+z³-3xyz =（x+y+z）（x²+y²+z²-xy-xz-yz）(这用到的是公式a³+b³+c³-3abc=(a²+b²+c²-ab-ac-bc)）
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)...
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...
x^3+y^3+z^3 -3xyz 这个怎么分解因式 各位老师帮帮忙
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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
一道难题,求解!
需要用到因式分解x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)分解过程参见：(x+y+z)^2-(x^2+y^2+z^2)=2xy+2xz+2yz 所以xy+xz+yz=[3*3-29]\/2=-10 所以 3xyz=x^3+y^3+z^3-(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=45-3*[29-(-10)]=-72 xyz=-...
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，...
