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: $225 + (9 \omega^2 + 64 \omega^4 + 48 \omega^3) + (

Vedclass · Accepted Answer

$248$

Vedclass · Answer

(D) Given expression: $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$ Expanding the squares: $225 + (9 \omega^2 + 64 \omega^4 + 48 \omega^3) + (9 \omega^4 + 64 \omega^2 + 48 \omega^3)$ Using the properties $\omega^3 = 1$ and $\omega^4 = \omega$: $225 + 9 \omega^2 + 64 \omega + 48 + 9 \omega + 64 \omega^2 + 48$ Grouping terms: $225 + 73(\omega^2 + \omega) + 96$ Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$: $225 + 73(-1) + 96$ Calculating the final value: $225 - 73 + 96 = 152 + 96 = 248$

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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