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(A) Given that $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$. These roots are the cube roots of unity,specifically $\omega$ and $\omega^2$. Let $\

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$x^2+x+1=0$

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(A) Given that $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$. These roots are the cube roots of unity,specifically $\omega$ and $\omega^2$. Let $\alpha = \omega = e^{i \frac{2\pi}{3}}$ and $\beta = \omega^2 = e^{i \frac{4\pi}{3}}$. We need to find the equation with roots $\alpha^{2023}$ and $\beta^{1012}$. First,calculate $\alpha^{2023} = (\omega)^{2023} = \omega^{2023 \pmod 3} = \omega^{2022+1} = \omega^1 = \alpha$. Next,calculate $\beta^{1012} = (\omega^2)^{1012} = \omega^{2024} = \omega^{2022+2} = \omega^2 = \beta$. Since the roots of the new equation are $\alpha$ and $\beta$,the required quadratic equation is the same as the original equation: $x^2+x+1=0$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2+x+1=0$,then the quadratic equation whose roots are $\alpha^{2023}$ and $\beta^{1012}$ is

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