分解因式 (x+y+z)的5次方-x的5次方-y的5次方-z的5次方
1个回答
2013-11-19
展开全部
因式分解=[(x+y+z)^5-x^5]-(y^5+z^5)
=(x+y+z-x)[(x+y+z)^4+x(x+y+z)^3+
x^2(x+y+z)^2+x^3(x+y+z)+x^4]-
(y+z)(y^4-y^3z+y^2z^2-yz^3+z^4)
=(y+z)[(x+y+z)^4+x(x+y+z)^3+x^2(x+y+z)^2+
x^3(x+y+z)+x^4-y^4+y^3z-y^2z^2+yz^3-z^4]
=(x+y+z-x)[(x+y+z)^4+x(x+y+z)^3+
x^2(x+y+z)^2+x^3(x+y+z)+x^4]-
(y+z)(y^4-y^3z+y^2z^2-yz^3+z^4)
=(y+z)[(x+y+z)^4+x(x+y+z)^3+x^2(x+y+z)^2+
x^3(x+y+z)+x^4-y^4+y^3z-y^2z^2+yz^3-z^4]
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