If omega is a complex cube root of unity,then | vedclass

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(C) We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Given expression: $(1 - \omega)(1 - \omega^2)(1 - \omega^4)(1 - \omega^8)$.Since $\ome

Vedclass · Accepted Answer

$9$

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(C) We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Given expression: $(1 - \omega)(1 - \omega^2)(1 - \omega^4)(1 - \omega^8)$.Since $\omega^3 = 1$,we have $\omega^4 = \omega$ and $\omega^8 = \omega^2$.Substituting these values,the expression becomes: $(1 - \omega)(1 - \omega^2)(1 - \omega)(1 - \omega^2) = [(1 - \omega)(1 - \omega^2)]^2$.Expanding the inner product: $(1 - \omega - \omega^2 + \omega^3) = (1 - (\omega + \omega^2) + 1) = (1 - (-1) + 1) = 3$.Thus,the expression is $3^2 = 9$.

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

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