If omega = frac{-1 + sqrt{3}i}{2},then (3 + o | vedclass

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(C) Given that $\omega = \frac{-1 + \sqrt{3}i}{2}$ is the complex cube root of unity,we know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omeg

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$16\omega$

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(C) Given that $\omega = \frac{-1 + \sqrt{3}i}{2}$ is the complex cube root of unity,we know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega^2 = -\omega$. Substituting this into the expression: $(3 + \omega + 3\omega^2)^4 = [3(1 + \omega^2) + \omega]^4$ $= [3(-\omega) + \omega]^4$ $= [-3\omega + \omega]^4$ $= [-2\omega]^4$ $= 16\omega^4$ Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 	imes \omega = 1 	imes \omega = \omega$. Therefore,$16\omega^4 = 16\omega$.

If $\omega = \frac{-1 + \sqrt{3}i}{2}$,then $(3 + \omega + 3\omega^2)^4 = $

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