x^2-y^2=(x+y)(x-y),x^5-y^5=?

Question

x^2-y^2=(x+y)(x-y),x^5-y^5=?

Answer 1

x^2-y^2=(x+y)(x-y),
x^5-y^5=?
设x^5-y^5=(x-y)(x^4+ax^3y+bx^2y^2+cxy^3+dy^4)=
=x^5+ax^4y+bx^3y^2+cx^2y^3+dxy^4-(x^4y+ax^3y^2+bx^2y^3+cxy^4+dy^5)=
=x^5+(a-1)x^4y+(b-a)x^3y^2+(c-b)x^2y^3+(d-c)xy^4-dy^5,
比较,得：
a-1=0,
b-a=0,
c-b=0,
d-c=0,
d=1,
解得：a=1,b=a=1,c=b=1,d=c=1,
所以,x^5-y^5=x^5-dy^5,又返回到起点!可见这种思路是不妥的.
设
x^5-y^5=(x^2-y^2)(x^3+ax^2y+bxy^2+y^3)=
=x^5+ax^4y+bx^3y^2+x^2y^3-(x^3y^2+ax^2y^3+bxy^4+y^5)=
=x^5+ax^4y+(b-1)x^3y^2+(1-a)x^2y^3-bxy^4-y^5,
a=0,b=1,a=1,b=0,
在一个式子里,a和b不能固定却要是两个数值,说明这个因式分解是不成功的!
以上是我的思路.先解到这里吧!抛出来-抛砖引玉吧!
以上回答于 回答者：我是杜鹃wsdj - 九级 2010-10-6 08:02
以下的思路和处理过程,可以完成x^5-y^5的因式分解!
x^5-y^5=
=x^2x^3-y^2y^3=
=x^2y^3{(x/y)^3-(x/y)+(x/y)-(y/x)^2}=
=x^2y^3{(x/y)[(x/y)^2-1]+(x/y)[1-(y/x)^3]}=
=x^2y^3(x/y)[(x/y+1)(x/y-1)+(1-y/x)(1+x/y+x^2/y^2)]=
=x^3y^2[(x/y+1)(x/y-1)+(1-y/x)(1+x/y+x^2/y^2)].
到此为止可以说也算完成了因式分解.
试着把分母去除、整式化：
^5-y^5=
=x^3y^2[(x/y+1)(x/y-1)+(1-y/x)(1+x/y+x^2/y^2)]=
=x^2[xy(x^2+xy)(x-y)+(x-y)(y^2+xy+x^2)]=
=x^2(x-y)[x^2y(x+y)+x^2+xy+y^2].
这个因式分解的结果看起来似乎更好些!