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(B) Given,$\omega^{3}=1$ and $1+\omega+\omega^{2}=0$.The determinant is $\Delta = \left|\begin{array}{ccc} 1 & \omega^{2} & 1-\omega \ \omega & 1 & 1

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$\omega^{2}-4$

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(B) Given,$\omega^{3}=1$ and $1+\omega+\omega^{2}=0$.The determinant is $\Delta = \left|\begin{array}{ccc} 1 & \omega^{2} & 1-\omega \ \omega & 1 & 1+\omega^{2} \ 1 & \omega & \omega^{2} \end{array}ight|$.Using $1+\omega^{2} = -\omega$,we get $\Delta = \left|\begin{array}{ccc} 1 & \omega^{2} & 1-\omega \ \omega & 1 & -\omega \ 1 & \omega & \omega^{2} \end{array}ight|$.Expanding along $R_{1}$:$\Delta = 1(\omega^{2} - (-\omega^{2})) - \omega^{2}(\omega^{3} - (-\omega)) + (1-\omega)(\omega^{2} - 1)$$\Delta = 1(2\omega^{2}) - \omega^{2}(1+\omega) + (\omega^{2} - 1 - \omega^{3} + \omega)$$\Delta = 2\omega^{2} - (\omega^{2} + \omega^{3}) + (\omega^{2} - 1 - 1 + \omega)$$\Delta = 2\omega^{2} - \omega^{2} - 1 + \omega^{2} - 2 + \omega$$\Delta = 2\omega^{2} + \omega - 3$.Since $1+\omega+\omega^{2}=0$,we have $\omega = -1-\omega^{2}$.$\Delta = 2\omega^{2} + (-1-\omega^{2}) - 3 = \omega^{2} - 4$.

If $\omega$ is an imaginary cube root of unity,then the value of $\left[\begin{array}{ccc}1 & \omega^{2} & 1-\omega^{4} \\ \omega & 1 & 1+\omega^{5} \\ 1 & \omega & \omega^{2}\end{array}\right]$ is

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