Question

If $1,\omega$ and ${\omega }^2$ are cube roots of unity, show that $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)$ upto $2$n factors $=1$.

Answer 1

Given,

$(1+w)(1+w^2)(1+w^4)(1+w^8)$ upto 2n terms

considering first 2 terms, we get,

$(1+w)(1+w^2)=1+w^2+w+w^3=0+w^3=1$

similarly, considering second 2 terms, we get,

$(1+w^4)(1+w^8)=(1+w)(1+w^2)=1$

It continues for all the remaining pair of terms,

therefore the product upto 2n terms equals 1.