x^3+y^3+z^3-3xyz因式分解
3个回答
展开全部
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)
不懂还可问，满意请及时采纳！o(∩_∩)o
= (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
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
x³+y³+z³-3xyz
=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)
=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
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
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)
= (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)
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
更多回答（1）
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
