Question

chưng minh rằng: X^8n+X^4n+1 chia hết cho X^2n+X^n+1 với mọi số tự nhiên n

 

Answer 1

\(\text{Ta có :}\)

\(x^{8n}+x^{4n}+1=x^{8n}+2x^{4n}+1-x^{4n}\)

\(=\left(x^{4n}+1\right)^2-\left(x^{2n}\right)^2\)

\(=\left(x^{4n}-x^{2n}+1\right)\left(x^{4n}+x^{2n}+1\right)\)

\(\text{Ta lại có :}\)

\(x^{4n}+x^{2n}+1=x^{4n}+2x^{2n}+1-x^{2n}\)

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

\(\Rightarrow x^{8n}+x^{4n}+1=\left(x^{4n}-x^{2n}+1\right)\left(x^{2n}-x^n+1\right)\left(x^{2n}+x^n+1\right)\)

\(\Rightarrow x^{8n}+x^{4n}+1⋮x^{2n}+x^n+1\)