Ta có công thức ${{x}^{k}}-1\,\,\,\vdots \,\,\,x-1$

Ta có hằng đẳng thức: ${{x}^{3}}-1=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$

Nên nếu đa thức $\vdots \,\,\,{{x}^{3}}-1$  thì nó cũng $\vdots \,\,\,{{x}^{2}}+x+1$

Câu 1: Chứng minh ${{x}^{2021}}+{{x}^{2020}}+1$ cho hết cho ${{x}^{2}}+x+1$

Xét $\left( {{x}^{2021}}+{{x}^{2020}}+1 \right)-\left( {{x}^{2}}+x+1 \right)$

$=\left( {{x}^{2021}}-{{x}^{2}} \right)+\left( {{x}^{2020}}-x \right)$

$={{x}^{2}}\left( {{x}^{2019}}-1 \right)+x\left( {{x}^{2019}}-1 \right)$

$=\left( {{x}^{2019}}-1 \right)\left( {{x}^{2}}+x \right)$

Vì ${{x}^{2019}}-1={{\left( {{x}^{673}} \right)}^{3}}-1\,\,\,\vdots \,\,\,\left( {{x}^{3}}-1 \right)$

$\Rightarrow \left( {{x}^{2019}}-1 \right)\left( {{x}^{2}}+x \right)\,\,\,\vdots \,\,\,\left( {{x}^{3}}-1 \right)$

$\Rightarrow \left( {{x}^{2019}}-1 \right)\left( {{x}^{2}}+x \right)\,\,\,\vdots \,\,\,\left( {{x}^{2}}+x+1 \right)$

$\Rightarrow \left( {{x}^{2021}}+{{x}^{2020}}+1 \right)-\left( {{x}^{2}}+x+1 \right)\,\,\,\vdots \,\,\,\left( {{x}^{2}}+x+1 \right)$

$\Rightarrow {{x}^{2021}}+{{x}^{2020}}+1\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

Câu 2: Chứng minh ${{x}^{3k+1}}+{{x}^{2}}+1\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

Có: ${{x}^{3k+1}}+{{x}^{2}}+1$

$={{x}^{3k+1}}-x+\left( {{x}^{2}}+x+1 \right)$

$=x\left( {{x}^{3k}}-1 \right)+\left( {{x}^{2}}+x+1 \right)$

Vì ${{x}^{3k}}-1\,\,\,\vdots \,\,\,{{x}^{3}}-1$

$\Rightarrow {{x}^{3k}}-1\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

$\Rightarrow x\left( {{x}^{3k}}-1 \right)+\left( {{x}^{2}}+x+1 \right)\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

$\Rightarrow {{x}^{3k+1}}+{{x}^{2}}+1\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

Câu 3: Chứng minh ${{x}^{3k+2}}+x+1\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

Có ${{x}^{3k+2}}+x+1$

$={{x}^{3k+2}}-{{x}^{2}}+\left( {{x}^{2}}+x+1 \right)$

$={{x}^{2}}\left( {{x}^{3k}}-1 \right)+\left( {{x}^{2}}+x+1 \right)$

Vì ${{x}^{3k-1}}\,\,\,\vdots \,\,\,{{x}^{3}}-1$

$\Rightarrow {{x}^{3k-1}}\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

$\Rightarrow {{x}^{2}}\left( {{x}^{3k-1}} \right)+\left( {{x}^{2}}+x+1 \right)\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

$\Rightarrow {{x}^{3k+2}}+x+1\,\,\,\vdots \,\,\,{{x}^{2}}+x+1$

Câu 4: Chứng minh ${{x}^{6}}-1\,\,\,\vdots \,\,\,{{x}^{4}}+{{x}^{2}}+1$

Xét ${{x}^{6}}-1+\left( {{x}^{4}}+{{x}^{2}}+1 \right)$

$={{x}^{6}}+{{x}^{4}}+{{x}^{2}}$

$={{x}^{2}}\left( {{x}^{4}}+{{x}^{2}}+1 \right)\,\,\,\vdots \,\,\,{{x}^{4}}+{{x}^{2}}+1$

$\Rightarrow {{x}^{6}}-1+\left( {{x}^{4}}+{{x}^{2}}+1 \right)\,\,\,\vdots \,\,\,{{x}^{4}}+{{x}^{2}}+1$

$\Rightarrow {{x}^{6}}-1\,\,\,\vdots \,\,\,{{x}^{4}}+{{x}^{2}}+1$