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

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(C) We know that $\omega^3 = 1$ and $\frac{1}{\omega} = \omega^2$,$\frac{1}{\omega^2} = \omega$. For $k=1$,$(\omega + \omega^2)^2 = (-1)^2 = 1$. For $

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$12$

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(C) We know that $\omega^3 = 1$ and $\frac{1}{\omega} = \omega^2$,$\frac{1}{\omega^2} = \omega$. For $k=1$,$(\omega + \omega^2)^2 = (-1)^2 = 1$. For $k=2$,$(\omega^2 + \omega)^2 = (-1)^2 = 1$. For $k=3$,$(\omega^3 + \frac{1}{\omega^3})^2 = (1 + 1)^2 = 4$. For $k=4$,$(\omega^4 + \frac{1}{\omega^4})^2 = (\omega + \omega^2)^2 = (-1)^2 = 1$. For $k=5$,$(\omega^5 + \frac{1}{\omega^5})^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$. For $k=6$,$(\omega^6 + \frac{1}{\omega^6})^2 = (1 + 1)^2 = 4$. Summing these values: $1 + 1 + 4 + 1 + 1 + 4 = 12$.

If $\omega$ is a complex cube root of unity,then $\sum_{k=1}^6\left(\omega^k+\frac{1}{\omega^k}\right)^2=$

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