If alpha is a non-real root of x^6=1,then fra | vedclass

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(C) Given that $x^6 = 1$,we have $x^6 - 1 = 0$. This can be factored as $(x^2-1)(x^4+x^2+1) = 0$ or $(x-1)(x^5+x^4+x^3+x^2+x+1) = 0$. Since $\alpha$ i

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$-\alpha^2$

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(C) Given that $x^6 = 1$,we have $x^6 - 1 = 0$. This can be factored as $(x^2-1)(x^4+x^2+1) = 0$ or $(x-1)(x^5+x^4+x^3+x^2+x+1) = 0$. Since $\alpha$ is a non-real root of $x^6=1$,it satisfies the equation $x^5+x^4+x^3+x^2+x+1=0$. Therefore,$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$. We can rearrange the terms as: $\alpha^5+\alpha^3+\alpha+1 = -(\alpha^4+\alpha^2)$ $\alpha^5+\alpha^3+\alpha+1 = -\alpha^2(\alpha^2+1)$ Dividing both sides by $(\alpha^2+1)$,we get: $\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1} = -\alpha^2$.

If $\alpha$ is a non-real root of $x^6=1$,then $\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}$ is equal to

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