题目

(x+y+z)5-x5-y5-z5,5都为次方,作因式分解,
有人说答案是5*(x + y)*(x + z)*(y + z)*(x^2 + y*x + z*x + y^2 + z^2 + y*z)，

提问时间：2020-10-08

答案

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]