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

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(A) We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$.Thus,$\frac{1-\sqrt{3}i}{2} = -\omega^2$ and $\frac{1+\sqrt

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$-2$

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(A) We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$.Thus,$\frac{1-\sqrt{3}i}{2} = -\omega^2$ and $\frac{1+\sqrt{3}i}{2} = -\omega$.Then,$\left(-\omega^2ight)^{2020} + (-\omega)^{2026} = \omega^{4040} + \omega^{2026} = \omega^2 + \omega = -1$.Next,consider the sum $\sum_{j=1}^6 (j+\omega)(j+\omega^2) = \sum_{j=1}^6 (j^2 + j(\omega+\omega^2) + \omega^3) = \sum_{j=1}^6 (j^2 - j + 1)$.Using summation formulas: $\sum_{j=1}^6 j^2 = \frac{6(7)(13)}{6} = 91$,$\sum_{j=1}^6 j = \frac{6(7)}{2} = 21$,and $\sum_{j=1}^6 1 = 6$.So,the sum is $91 - 21 + 6 = 76$.The sine term becomes $\sin\left(76 	imes \frac{3\pi}{152}ight) = \sin\left(\frac{3\pi}{2}ight) = -1$.Finally,the total expression is $-1 + (-1) = -2$.

If $\omega$ is a complex cube root of unity,then $\left(\frac{1-\sqrt{3} i}{2}\right)^{2020}+\left(\frac{1+\sqrt{3} i}{2}\right)^{2026} +\sin \left(\sum_{j=1}^6(j+\omega)(j+\omega^2) \frac{3 \pi}{152}\right)=$

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