Question

If $1,\omega ,{\omega }^2$ be the three cube roots of unity, then
$(1+\omega )\prod _{n=1}^{2n-1}(1+{\omega }^{2^n})=$

Answer 1

Given $(1+\omega )\prod _{n=1}^{2n-1}(1+{\omega }^{2^n})$
Expanding, we get $(1+\omega )(1+{\omega }^2)(1+{\omega }^4)(1+{\omega }^8)\cdots (1+{\omega }^{2^{2n-1}})$ to $2n$ factors
$(1+\omega )(1+{\omega }^2)(1+\omega )(1+{\omega }^2)\cdots \cdots (1+{\omega }^2)$ upto $2n$ factors
Using property $1+\omega +{\omega }^2=0$
$=(-{\omega }^2)(-\omega )(-{\omega }^2)(-\omega )\cdots 2n\text{terms}$
$=1[\because {\omega }^3=1]$