If omega is a non-real cube root of unity and | vedclass

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(A) Given,$x = \omega^2 - \omega - 3$. Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$. Substituting this,$x = (-1 - \omega) - \ome

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$5$

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(A) Given,$x = \omega^2 - \omega - 3$. Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$. Substituting this,$x = (-1 - \omega) - \omega - 3 = -2\omega - 4$. Alternatively,using $\omega^2 - \omega - 1 = 0$ is not correct,but $\omega^2 + \omega + 1 = 0$ is. Let us re-evaluate: $x = \omega^2 - \omega - 3$. Since $\omega^2 = -1 - \omega$,$x = -1 - \omega - \omega - 3 = -2\omega - 4$. Then $x + 4 = -2\omega$. Squaring both sides,$(x + 4)^2 = 4\omega^2$. $x^2 + 8x + 16 = 4(-1 - \omega) = -4 - 4\omega$. Since $-4\omega = 2(x + 4)$,we have $x^2 + 8x + 16 = -4 + 2x + 8$. $x^2 + 6x + 12 = 0$. Now,perform polynomial division of $x^4 + 6x^3 + 10x^2 - 12x - 19$ by $x^2 + 6x + 12$. $x^4 + 6x^3 + 10x^2 - 12x - 19 = (x^2 + 6x + 12)(x^2 - 2) + 5$. Since $x^2 + 6x + 12 = 0$,the expression equals $0(x^2 - 2) + 5 = 5$.

If $\omega$ is a non-real cube root of unity and $x = \omega^2 - \omega - 3$,then the value of $x^4 + 6x^3 + 10x^2 - 12x - 19$ is

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