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(A) Given that $\omega$ is a complex cube root of unity,we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Since $x = \omega^2 - \omega + 2$,we ca

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$x^2 - 4x + 7 = 0$

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(A) Given that $\omega$ is a complex cube root of unity,we have $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Since $x = \omega^2 - \omega + 2$,we can substitute $\omega^2 = -1 - \omega$ into the expression:$x = (-1 - \omega) - \omega + 2 = 1 - 2\omega$.Thus,$2\omega = 1 - x$,which implies $\omega = \frac{1 - x}{2}$.Substitute this into the equation $1 + \omega + \omega^2 = 0$:$1 + \left(\frac{1 - x}{2}\right) + \left(\frac{1 - x}{2}\right)^2 = 0$.Multiply by $4$ to clear the denominator:$4 + 2(1 - x) + (1 - x)^2 = 0$.$4 + 2 - 2x + 1 - 2x + x^2 = 0$.$x^2 - 4x + 7 = 0$.

If $\omega$ is a complex cube root of unity and $x = \omega^2 - \omega + 2$,then:

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