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

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(D) Given expression: $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$Expanding the squares using $(a+b)^2 = a^2+2ab+b^2$ and using $\omega^3 = 1

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$248$

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(D) Given expression: $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$Expanding the squares using $(a+b)^2 = a^2+2ab+b^2$ and using $\omega^3 = 1$ and $\omega^4 = \omega$:$= 225 + (9 \omega^2 + 48 \omega^3 + 64 \omega^4) + (9 \omega^4 + 48 \omega^3 + 64 \omega^2)$$= 225 + (9 \omega^2 + 48(1) + 64 \omega) + (9 \omega + 48(1) + 64 \omega^2)$Grouping the terms:$= 225 + 48 + 48 + (9+64) \omega^2 + (64+9) \omega$$= 225 + 96 + 73(\omega^2 + \omega)$Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$:$= 321 + 73(-1)$$= 321 - 73 = 248$

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

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