Question

If $\omega$ is an imaginary cube root of unity, then $(1-{\omega }^4)(1-{\omega }^8)(1-{\omega }^{22})(1-{\omega }^{44})$ equals to

• A. ${\omega }^2$
• B. $\omega$
• C. $9$
• D. $0$

Answer 1

Answer: (C)

$9$

We, know that ${\omega }^3=1$ And, $1+\omega +{\omega }^2=0$

According to question,

$(1-{\omega }^4)(1-{\omega }^8)(1-{\omega }^{22})(1-{\omega }^{44})$

$⟹(1-{\omega }^3.\omega )(1-{\omega }^6.{\omega }^2)(1-{\omega }^{21}.\omega )(1-{\omega }^{42}.{\omega }^2)$

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

$⟹(1-\omega {)}^2(1-{\omega }^2{)}^2$

$⟹(1-\omega {)}^4(1+\omega {)}^2$

$⟹(1-\omega {)}^4(-{\omega }^2{)}^2$

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

$⟹(-3\omega {)}^2(\omega )$

$⟹9{\omega }^3$

$⟹9$