2)Ta có: \(x^{3m+1}+x^{3n+2}+1\)= \(x^{3m+1}-x+x^{3n+2}-x^2+x^2+x+1\)
= \(x\left(x^{3m}-1\right)+x^2\left(x^{3n}-1\right)+\left(x^2+x+1\right)\)
Ta thấy: \(x^{3m}-1=\left(x^3\right)^m-1=\left(x^3-1\right)k\) \(⋮\) \(x^3-1\)
\(x^{3n}-1=\left(x^3\right)^n-1=\left(x^3-1\right)h\) \(⋮\) \(x^3-1\)
Do đó: \(x\left(x^{3m}-1\right)+x^2\left(x^{3n}-1\right)+\left(x^2+x+1\right)\) chia hết cho \(x^2+x+1\)
Vậy \(x^{3m+1}+x^{3n+2}+1\) chia hết cho \(x^2+x+1\)