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

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(D) Given expression: $225 + (3\omega + 8\omega^2)^2 + (3\omega^2 + 8\omega)^2$ Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$. Su

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$248$

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(D) Given expression: $225 + (3\omega + 8\omega^2)^2 + (3\omega^2 + 8\omega)^2$ Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$. Substitute $\omega^2$ in the terms: $(3\omega + 8\omega^2) = 3\omega + 8(-1 - \omega) = 3\omega - 8 - 8\omega = -5\omega - 8$. $(3\omega^2 + 8\omega) = 3(-1 - \omega) + 8\omega = -3 - 3\omega + 8\omega = 5\omega - 3$. Now,expand the squares: $(-5\omega - 8)^2 = 25\omega^2 + 80\omega + 64$. $(5\omega - 3)^2 = 25\omega^2 - 30\omega + 9$. Summing these: $25\omega^2 + 80\omega + 64 + 25\omega^2 - 30\omega + 9 = 50\omega^2 + 50\omega + 73$. Using $50(\omega^2 + \omega) = 50(-1) = -50$: Result $= 225 - 50 + 73 = 248$.

If $\omega$ is a complex cube root of unity,then $225 + (3\omega + 8\omega^2)^2 + (3\omega^2 + 8\omega)^2 = $

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