因式分解x^5-y^5

因式分解x^5-y^5
因式分解x^5-y^5

因式分解x^5-y^5
x^5-y^5
=(x^3+y^3)(x^2-y^2)+x^3y^2-x^2y^3
=(x+y)(x^2-xy+y^2)(x+y)(x-y)+x^2y^2(x-y)
=(x-y)[(x+y)^2(x^2-xy+y^2)+x^2y^2]
=(x-y)[(x+y)(x^3+y^3)+x^2y^2)
=(x-y)[x^4+x^3y+x^2y^2+xy^3+y^4)

x^5-y^5
=x^5-x^4y+x^4y-x^3y^2+x^3y^2-x^2y^3+x^2y^3-xy^4+xy^4-y^5
=x^4(x-y)+x^3y(x-y)+x^2y^2(x-y)+xy^3(x-y)+y^4(x-y)
=(x-y)(x^4x^3y+x^2y^2+xy^3+y^4)

X*X*X*X*X-Y*Y*Y*Y*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：...

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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].
采纳吧

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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+...

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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！

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