Question

If $1,\omega ,{\omega }^2$ are the three cube roots of unity,$(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)$...to 2n factors $=$

• A. $2^{3n}$
• B. $2^{2n}$
• C. $2^n$
• D. none of these

Answer 1

Answer: (B)

$2^{2n}$

$(1-w+w^2)=(1+w+w^2-2w)=(-2w)$
Similarly we get,
$(1-w^2+w^4)=(1+w+w^2-2w^2)=(-2w^2)$
The third factor will be,
$(1-w^4+w^8)=(1+w+w^2-2w)=(-2w)$
Thus we get $(-2w)and(-2w^2)$ as succesive factors,
and we have $2n$ factors,so the product becomes,
$(-2w)(-2w^2)(-2w)(-2w^2).........(-2w^2)$
$={(-2w)}^n{(-2w^2)}^n$
$={\left(2\right)}^{2n}{{(w}^3)}^{2n}$
$={\left(2\right)}^{2n}$