已知x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
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(1)x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
在上式中取y为-y,则有
x^5-y^5
=x^5+(-y)^5
=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
(2)
x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
在上式中分别取y=-1和1，
则有
x^5-1
=x^5+(-1)
=x^5+(-1)^5
=(x-1)(x^4+x^3+x^2+x+1)
x^5+1
=x^5+1^5
=(x+1)(x^4-x^3+x^2-x+1)
从而
x^10-1
=(x^5)^2-1^2
=(x^5-1)(x^5+1)
=(x-1)(x^4+x^3+x^2+x+1)(x+1)(x^4-x^3+x^2-x+1)
=(x-1)(x+1)(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)
在上式中取y为-y,则有
x^5-y^5
=x^5+(-y)^5
=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
(2)
x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
在上式中分别取y=-1和1，
则有
x^5-1
=x^5+(-1)
=x^5+(-1)^5
=(x-1)(x^4+x^3+x^2+x+1)
x^5+1
=x^5+1^5
=(x+1)(x^4-x^3+x^2-x+1)
从而
x^10-1
=(x^5)^2-1^2
=(x^5-1)(x^5+1)
=(x-1)(x^4+x^3+x^2+x+1)(x+1)(x^4-x^3+x^2-x+1)
=(x-1)(x+1)(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)
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