Question

Phân tích đa thức thành nhân tử 

x\(^{10}\)+x\(^2\)+1

Answer 1

\(x^{10}+x^2+1\)

\(=x^{10}-x^8+x^4+x^8-x^6+x^2+x^6-x^4+1\)

\(=x^4\left(x^6-x^4+1\right)+x^2\left(x^6-x^4+1\right)+\left(x^6-x^4+1\right)\)

\(=\left(x^6-x^4+1\right)\left(x^4+x^2+1\right)\)

\(=\left(x^6-x^4+1\right)\left[x^4+2x^2+1-x^2\right]\)

\(=\left(x^6-x^4+1\right)\left[\left(x^2+1\right)^2-x^2\right]\)

\(=\left(x^6-x^4+1\right)\left(x^2+1+x\right)\left(x^2+1-x\right)\)