If omega neq 1 is a cube root of unity,then t | vedclass

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(D) Given that $\omega$ is a cube root of unity,we know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Simplifying the powers of $\omega$ in the

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$0$

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(D) Given that $\omega$ is a cube root of unity,we know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Simplifying the powers of $\omega$ in the determinant:$\omega^9 = (\omega^3)^3 = 1^3 = 1$$\omega^{27} = (\omega^3)^9 = 1^9 = 1$$\omega^{31} = \omega^{30} \cdot \omega = (\omega^3)^{10} \cdot \omega = 1^{10} \cdot \omega = \omega$$\omega^{17} = \omega^{15} \cdot \omega^2 = (\omega^3)^5 \cdot \omega^2 = 1^5 \cdot \omega^2 = \omega^2$$\omega^{30} = (\omega^3)^{10} = 1^{10} = 1$$\omega^{41} = \omega^{39} \cdot \omega^2 = (\omega^3)^{13} \cdot \omega^2 = 1^{13} \cdot \omega^2 = \omega^2$$\omega^{19} = \omega^{18} \cdot \omega = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$Substituting these values into the determinant:$\Delta = \left|\begin{array}{ccc}\omega+\omega^2 & \omega^2+1 & 1+\omega \ 1+\omega & \omega+\omega^2 & \omega^2+1 \ 1+\omega^2 & \omega^2+\omega & \omega+1\end{array}ight|$Since $1+\omega+\omega^2 = 0$,we have $\omega+\omega^2 = -1$,$\omega^2+1 = -\omega$,and $1+\omega = -\omega^2$.$\Delta = \left|\begin{array}{ccc}-1 & -\omega & -\omega^2 \ -\omega^2 & -1 & -\omega \ -\omega & -\omega^2 & -1\end{array}ight|$Taking $-1$ common from each row:$\Delta = (-1)^3 \left|\begin{array}{ccc}1 & \omega & \omega^2 \ \omega^2 & 1 & \omega \ \omega & \omega^2 & 1\end{array}ight|$Using the property of cyclic determinants,this determinant is equal to $1(1-\omega^3) - \omega(\omega^2-\omega^2) + \omega^2(\omega^4-\omega) = 1(1-1) - 0 + \omega^2(\omega-\omega) = 0$.Thus,the value is $0$.

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