Question

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

• A. $2^n$
• B. $2^{2n}$
• C. $2^{4n}$
• D. None of these

Answer 1

The correct option is B $2^{2n}$

We have $(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)(1-{\omega }^8+{\omega }^{16})...$ to $2n$ factors

$=(1-\omega +{\omega }^2)(1-{\omega }^2+\omega )(1-\omega +{\omega }^2)(1-{\omega }^2+\omega )...$ to $2n$ factors

$[\because {\omega }^4=w,{\omega }^8={\omega }^2,{\omega }^{16}=\omega ,...]$

$=(-2\omega )(-2{\omega }^2)(-2\omega )(-2{\omega }^2)...$ to $2n$ factors

$=\left(2^2{\omega }^3\right)\left(2^2{\omega }^3\right)...$ to $n$ factors $[\because (-2\omega )(-2{\omega }^2)=2^2{\omega }^3=2^2]$

$={\left(2^2\right)}^n=2^{2n}$.