If alpha and beta are non-real roots of x^3-x | vedclass

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(C) Given equation is $x^3-x^2-x-2=0$. By testing $x=2$,we get $8-4-2-2=0$,so $(x-2)$ is a factor. Dividing by $(x-2)$,we get $(x-2)(x^2+x+1)=0$. The

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$1+\alpha+\beta$

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(C) Given equation is $x^3-x^2-x-2=0$. By testing $x=2$,we get $8-4-2-2=0$,so $(x-2)$ is a factor. Dividing by $(x-2)$,we get $(x-2)(x^2+x+1)=0$. The non-real roots $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$. These are the complex cube roots of unity,$\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$. We need to evaluate $\alpha^{2020}+\beta^{2020}+\alpha^{2020} \cdot \beta^{2020}$. Since $\alpha=\omega$ and $\beta=\omega^2$,we have $\alpha^{2020}=\omega^{2020}=\omega^{3 	imes 673 + 1}=\omega$ and $\beta^{2020}=(\omega^2)^{2020}=\omega^{4040}=\omega^{3 	imes 1346 + 2}=\omega^2$. Thus,$\alpha^{2020}+\beta^{2020}+\alpha^{2020} \cdot \beta^{2020} = \omega + \omega^2 + \omega \cdot \omega^2 = \omega + \omega^2 + \omega^3 = \omega + \omega^2 + 1 = 0$. However,looking at the options,$1+\alpha+\beta = 1 + \omega + \omega^2 = 0$. Therefore,the expression equals $0$,which is equivalent to $1+\alpha+\beta$.

If $\alpha$ and $\beta$ are non-real roots of $x^3-x^2-x-2=0$,then $\alpha^{2020}+\beta^{2020}+\alpha^{2020} \cdot \beta^{2020}=$

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