因式分解(x+y+z)^5-(x
解法一
利用公式
(a+b+c)^5-(a^5+b^5+c^5)
=(5/2)(a+b)(b+c)(c+a)[(a+b)^2+(b+c)^2+(c+a)^2]
取a=x+y-z,b=x-y+z,c=-x+y+z,则a+b+c=x+y+z
代入得:
(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5
=(5/2)(2x)(2y)(2z)[(2x)^2+(2y)^2+(2z)^2]
=80xyz(x^2+y^2+z^2)
解法二
令x=0,则
(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5
=(y+z)^5-(y-z)^5-(-y...全部
解法一
利用公式
(a+b+c)^5-(a^5+b^5+c^5)
=(5/2)(a+b)(b+c)(c+a)[(a+b)^2+(b+c)^2+(c+a)^2]
取a=x+y-z,b=x-y+z,c=-x+y+z,则a+b+c=x+y+z
代入得:
(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5
=(5/2)(2x)(2y)(2z)[(2x)^2+(2y)^2+(2z)^2]
=80xyz(x^2+y^2+z^2)
解法二
令x=0,则
(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5
=(y+z)^5-(y-z)^5-(-y+z)^5-(y+z)^5
=0
所以分解式中含因式x,同样也含y、z
从而分解式中含因式xyz
又因为(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5是五次轮换对称式,
所以除去xyz外,还有二次轮换对称式,
只能是k(x^2+y^2+z^2)
或者k[(x+y)^2+(y+z)^2+(z+x)^2]
或者k[(x-y)^2+(y-z)^2+(z-x)^2
用代定系数法可算出k=80,并排除后两种情况,
可得,原式=80xyz(x^2+y^2+z^2)
解法三
利用公式
a^5+b^5=(a+b)(a^4+a^3*b+a^2*b^2+a*b^3+b^4)
取a=x+y+z,b=x-y-z
则(x+y+z)^5-(-x+y+z)^5
=(x+y+z)^5+(x-y-z)^5
=2x[ ]
取c=x+y-z,d=x-y+z
则(x+y-z)^5+(x-y+z)^5
=2x[ ]
提取x,再找y、z
不想算了。
。收起