If w = frac{-1 + i sqrt{3}}{2},where i = sqrt | vedclass

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(C) Given $w = \frac{-1 + i \sqrt{3}}{2}$,which is the complex cube root of unity,denoted as $\omega$. We know that $1 + \omega + \omega^2 = 0$,which

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$16w$

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(C) Given $w = \frac{-1 + i \sqrt{3}}{2}$,which is the complex cube root of unity,denoted as $\omega$. We know that $1 + \omega + \omega^2 = 0$,which implies $\omega^2 = -1 - \omega$. We need to evaluate $(3 + \omega + 3 \omega^2)^4$. Substitute $\omega^2 = -1 - \omega$ into the expression: $(3 + \omega + 3(-1 - \omega))^4 = (3 + \omega - 3 - 3 \omega)^4$. $= (-2 \omega)^4$. $= 16 \omega^4$. Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$. Therefore,$16 \omega^4 = 16 \omega$.

If $w = \frac{-1 + i \sqrt{3}}{2}$,where $i = \sqrt{-1}$,then the value of $(3 + w + 3 w^2)^4$ is

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