设m,n,p为任意非负整数,证明x^2+x+1|x^3m+x^(3n+1)+x^(3p+2)

x^3m+x^(3n+1)+x^(3p+2)
=x^3m-1+x^(3n+1)-x+x^(3p+2)-x^2+1+x+x^2
=(x^3-1)(1+x^3+...+x^(3m-3)) + x(x^3-1)(1+x^3+...+x^(3n-3)) +x^2(x^3-1)(1+x^3+...+x^(3p-3))+1+x+x^2
=(x-1)(1+x+x^2)(1+x^3+...+x^(3m-3)) + x(x-1)(1+x+x^2)(1+x^3+...+x^(3n-3)) +x^2(x-1)(1+x+x^2)(1+x^3+...+x^(3p-3))+1+x+x^2
每一项都有因式1+x+x^2
因此,x^2+x+1|x^3m+x^(3n+1)+x^(3p+2)

证明什么，你题目是不是错了

证明：原式=x^3m+x^(3n+1)+x^(3p+2)=[x^3m-1]+[x^(3n+1)-x]+[x^(3p+2)-x^2]+(1+x+x^2)
=[x^3m-1]+x[x^3n-1]+x^2[x^3p-1]+(1+x+x^2)
由于：x^2+x+1|x^3m-1；x^2+x+1|x^3n-1；x^2+x+1|x^3p-1
所以：x^2+x+1|x^3m+x^(3n+1)+x^(3p+2)