Ta có:\(x^{3m+1}+x^{3n+2}+1=x^{3m}x-x+3^{3n}-x^2+x^2+x+1=x\left(\left(x^3\right)^m-1\right)+x^2\left(\left(x^3\right)^n-1\right)+\left(x^2+x+1\right)\)Ta lại có: (Hằng đẳng thức)
\(a^n+b^n=\left(a+b\right)\left(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}\right)\)chia hết cho a+b
=>\(\left(x^3\right)^m-1\)chia hết cho \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)chia hết cho \(x^2+x+1\)
và \(\left(x^3\right)^n-1\)chia hết cho \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)chia hết cho \(x^2+x+1\)
mà \(x^{3m+1}+x^{3n+2}+1=x^{3m}x-x+3^{3n}-x^2+x^2+x+1=x\left(\left(x^3\right)^m-1\right)+x^2\left(\left(x^3\right)^n-1\right)+\left(x^2+x+1\right)\)
=> \(x^{3m+1}+x^{3n+2}+1\) chia hết cho \(x^2+x+1\)
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Xét
\(x^{3m+1}+x^{3n+2}+1-\left(x^2+x+1\right)\)
\(=x^{3m}.x+x^{3n}.x^2+1-x^2-x-1\)
\(=x\left(x^{3m}-1\right)+x^2\left(x^{3n}-1\right)\)
Do \(x^{3m}-1=\left(x^3\right)^m-1^m⋮x^3-1⋮x^2+x+1\)
\(x^{3n}-1=\left(x^3\right)^n-1^n⋮x^3-1⋮x^2+x+1\)
\(\Rightarrow x\left(x^{3m}-1\right)+x^2\left(x^{3n}-1\right)⋮x^2+x+1\)
\(\Rightarrow x^{3m+1}+x^{3n+2}+1-\left(x^2+x+1\right)⋮x^2+x+1\)
\(\Rightarrow x^{3m+1}+x^{3n+2}+1⋮x^2+x+1\)
Ta có:\(x^{3m+1}+x^{3n+2}+1\)
\(=\left(x^{3m+1}-x\right)+\left(x^{3n+2}-x^2\right)+\left(x^2+x+1\right)\)
\(=x\left(3^{3m}-1\right)+x^2\left(3^{3n}-1\right)+\left(x^2+x+1\right)\)
Do \(\hept{\begin{cases}x^{3m}-1⋮x^3-1\\x^{3n}-1⋮x^3-1\end{cases}}\Rightarrow x\left(x^{3m}-1\right)+x^2\left(x^{3n}-1\right)⋮x^3-1\left(1\right)\)
Mặt khác:\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\Rightarrow x^3-1⋮x^2+x+1\left(2\right)\)
Từ (1);(2) suy ra điều cần chứng minh
