Question

If $\omega$ is cubic root of unity then $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)=$

Answer 1

we know that $w^3=\mathrm{1.......}\left[1\right]$

$w^3+w^2+1=\mathrm{0...............}\left[2\right]$

hence, using $1$ and $2$;

$(1+w)(1+w^2)(1+w^4)(1+w^8)=(1+w)(1+w^2)(1+w^{3}w)(1+(w^3{)}^2{w}^2)$

$=(1+w)(1+w^2)(1+w)(1+w^2)=\left[\right(1+w\left)\right(1+w^2){]}^2=[1+w+w^2+w^3{]}^2=[1+0{]}^2=1$