If omega is an imaginary cube root of unity,t | vedclass

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(A) Given that $\omega^{3} = 1$ and $1+\omega+\omega^{2} = 0$. We know that $1+\omega^{2} = -\omega$ and $1+\omega = -\omega^{2}$. Also,$\omega^{3} =

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$2^{2n}$

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(A) Given that $\omega^{3} = 1$ and $1+\omega+\omega^{2} = 0$. We know that $1+\omega^{2} = -\omega$ and $1+\omega = -\omega^{2}$. Also,$\omega^{3} = 1, \omega^{4} = \omega, \omega^{8} = \omega^{2}, \omega^{16} = \omega$. The expression is $P = (1-\omega+\omega^{2})(1-\omega^{2}+\omega)(1-\omega+\omega^{2})(1-\omega^{2}+\omega) \ldots$ ($2n$ factors). Substituting $1+\omega^{2} = -\omega$ and $1+\omega = -\omega^{2}$: $P = (-\omega-\omega)(-\omega^{2}-\omega^{2})(-\omega-\omega)(-\omega^{2}-\omega^{2}) \ldots$ ($2n$ factors). $P = (-2\omega)(-2\omega^{2})(-2\omega)(-2\omega^{2}) \ldots$ ($2n$ factors). There are $n$ pairs of $(-2\omega)(-2\omega^{2}) = 4\omega^{3} = 4(1) = 4$. Thus,$P = (4)^{n} = 2^{2n}$.

If $\omega$ is an imaginary cube root of unity,then the value of $(1-\omega+\omega^{2}) \cdot(1-\omega^{2}+\omega^{4}) \cdot(1-\omega^{4}+\omega^{8}) \cdot \ldots$ ($2n$ factors) is

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