If $1, \omega, \omega^2$ are the cube roots of unity, then $\Delta=\left|\begin{array}{ccc}1 & \omega^n & \omega^{2 n} \ \omega^n & \omega^{2 n} & 1 \ \omega^{2 n} & 1 & \omega^n\end{array} ight|$ is equal to

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If $1, \omega, \omega^2$ are the cube roots of unity, then $\Delta=\left|\begin{array}{ccc}1 & \omega^n & \omega^{2 n} \ \omega^n & \omega^{2 n} & 1 \ \omega^{2 n} & 1 & \omega^n\end{array}ight|$ is equal to

Aicharya · Accepted Answer

Applying $\mathrm{R}_1 ightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3$ As, $1+\omega^{\mathrm{n}}+\omega^{2 \mathrm{n}}=0 ; \quad 	herefore \Delta=0$

If $1, \omega, \omega^2$ are the cube roots of unity, then $\Delta=\left|\begin{array}{ccc}1 & \omega^n &…

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