Question

$(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)(1-{\omega }^8+{\omega }^{16})=$

• A. $4$
• B. $8$
• C. $12$
• D. $16$

Answer 1

The correct option is D $16$

We have to find the value of $(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)(1-{\omega }^8+{\omega }^{16})$$\to \left(1\right)$

We know that since 1,$\omega ,{\omega }^2$are cube roots of unity

$1+\omega +{\omega }^2=0$

And also

${\omega }^1={\omega }^4=\dots ={\omega }^{3n+1}$

${\omega }^2={\omega }^5=\dots ={\omega }^{3n+2}$

${\omega }^3={\omega }^6=\dots ={\omega }^{3n}$

$\therefore$ $1+{\omega }^2=-\omega$

Now

$(1-\omega +{\omega }^2)$

$=(1+{\omega }^2-\omega )$

$=-2\omega$ ($\because 1+{\omega }^2=-\omega )\to (2)$

Also

$(1-{\omega }^2+{\omega }^4)=-2{\omega }^2$ (\because ${\omega }^1={\omega }^4=\dots ={\omega }^{3n+1}$ &($\because 1+\omega =-{\omega }^2$)

For the third term

$(1-{\omega }^4+{\omega }^8)=(1-{\omega }^1+{\omega }^2)$

$=-2\omega \to \left(3\right)$

Similarly

Fourth term is

$(1-{\omega }^8+{\omega }^{16})=-2{\omega }^2\to \left(4\right)$

Substituting values of (2),(3)&(4) in (1)

$(-2\omega )(-2{\omega }^2)(-2\omega )(-2{\omega }^2)=16{\omega }^3{\omega }^3=16(\because {\omega }^3=1)$

Hence option (D) is the correct answer