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

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(D) Given the determinant equation: $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \ \omega & x + 1 + \omega^2 & 1 \ \omega^2 & 1 & x + 1 + \o

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$-1$

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(D) Given the determinant equation: $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \ \omega & x + 1 + \omega^2 & 1 \ \omega^2 & 1 & x + 1 + \omega \end{array} ight| = 0$. Applying the row operation $R_1 ightarrow R_1 + R_2 + R_3$: The sum of the first row becomes: $(x + 2 + \omega + \omega^2) = (x + 1 + (1 + \omega + \omega^2)) = (x + 1 + 0) = (x + 1)$. Similarly,the other elements in the first row become: $(\omega + x + 1 + \omega^2 + 1) = (x + 1 + (1 + \omega + \omega^2)) = (x + 1)$. $(\omega^2 + 1 + x + 1 + \omega) = (x + 1 + (1 + \omega + \omega^2)) = (x + 1)$. Factoring out $(x + 1)$ from the first row: $(x + 1) \left| \begin{array}{ccc} 1 & 1 & 1 \ \omega & x + 1 + \omega^2 & 1 \ \omega^2 & 1 & x + 1 + \omega \end{array} ight| = 0$. Since the determinant part evaluates to a non-zero value for $x 
eq -1$,the equation simplifies to $(x + 1)^3 = 0$. Therefore,the root is $x = -1$.

If $\omega$ is a cube root of unity,then the root of the equation $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \\ \omega & x + 1 + \omega^2 & 1 \\ \omega^2 & 1 & x + 1 + \omega \end{array} \right| = 0$ is:

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