Question

CMR (x^6 -1) chia het cho (x^2 -x +1)

Answer 1

\(x^3+1=\left(x+1\right)\left(x^2-x+1\right)\)

=> \(\left(x^6-1\right)=\left(\left(x^3\right)^2-1\right)=\left(x^3-1\right)\left(x^3+1\right)=\left(x^3-1\right)\left(x+1\right)\left(x^2-x+1\right)⋮x^2-x+1\)

Answer 2

Dạo này bận quá ít thời gian làm =((( 

\(x^6-1\)  

\(=\left(x^3\right)^2-1^2\) 

\(=\left(x^3-1\right)\left(x^3+1\right)\) 

\(=\left(x^3-1^3\right)\left(x^3+1^3\right)\) 

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)⋮\left(x^2-x+1\right)\forall x\left(đpcm\right)\)

Answer 3

\(x^6-1\)

\(=\left(x^3\right)^2-1^2\)

\(=\left(x^3-1\right)\left(x^3+1\right)\)

\(=\left(x^3-1\right)\left(x+1\right)\left(x^2-x+1\right)⋮x^2-x+1\)

\(\Rightarrowđpcm\)