# If $\omega$ is a cube root of unity then the value of $(1-\omega)(1-\omega^{2})(1-\omega^{4})(1-\omega^{8})$ is

$\begin{array}{1 1}(1)9&(2)-9\\(3)16&(4)32\end{array}$

## 1 Answer

$\omega$ is a cube root of unity
then $\omega^3=1$ and $1+ \omega+\omega^2=0$
$(1- \omega)(1- \omega^2)(1-\omega^4)(1- \omega^8)=0$
$(1- \omega)(1- \omega^2)(1-\omega^3. \omega)(1- \omega^3. \omega^3. \omega^2)$
$\qquad= (1-\omega)(1-\omega^2)(1- \omega)(1 - \omega^2)$
$\qquad= (1-\omega)(1-\omega)^2(1- \omega)(1 - \omega^2)^2$
$\qquad= [(1-\omega)(1-\omega^2)]^2$
$\qquad= [(1-\omega)(1-\omega^2- \omega + \omega^3)]^2$
$\qquad= [1-(\omega + \omega^2)+1]^2$
$\qquad= [1-(-1)+1]^2$
$\qquad= [1+1+1]^2$
$\qquad= 3^2=9$
Hence 1 is the correct answer.
answered May 15, 2014 by

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