Question

If $\omega$ is a complex cube root of unity then $(1-\omega +{\omega }^2)(1-{\omega }^2+{\omega }^4)(1-{\omega }^4+{\omega }^8)(1-{\omega }^8+{\omega }^{16})$ is

• A. $16$
• B. $17$
• C. $15$
• D. $14$

Answer 1

Answer: (A)

$16$

We have,

${\omega }^3=1⟹1+\omega +{\omega }^2=0\because \omega \ne 1$ ................. $\left(i\right)$

$⟹1+\omega +{\omega }^2-2\omega =-2\omega$ i.e. $1-\omega +{\omega }^2=-2\omega$ ...................... $\left(ii\right)$

From $\left(i\right)$

$1-{\omega }^2+{\omega }^4=1-{\omega }^2+\omega =-2{\omega }^2$ ............. $\left(iii\right)$

$1-{\omega }^4+{\omega }^8=1-\omega +{\omega }^2=-2\omega$ .............. From $\left(ii\right)$

$1-{\omega }^8+{\omega }^{16}=1-{\omega }^2+\omega =-2{\omega }^2$ ............ From $\left(iii\right)$

Using all above equalities, we get

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

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

$=2^4({\omega }^3.{\omega }^3)$

$\therefore 2{}^4=16$