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(A) The given equation is $1+x+x^2=0$. Multiplying by $(x-1)$,we get $(x-1)(1+x+x^2)=0$,which implies $x^3-1=0$,so $x^3=1$. The roots of $1+x+x^2=0$ a

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$0$

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(A) The given equation is $1+x+x^2=0$. Multiplying by $(x-1)$,we get $(x-1)(1+x+x^2)=0$,which implies $x^3-1=0$,so $x^3=1$. The roots of $1+x+x^2=0$ are the complex cube roots of unity,$\omega$ and $\omega^2$. Let $\alpha = \omega$ and $\beta = \omega^2$. We need to find the value of $\alpha^4+\beta^4+\alpha^{-4}\beta^{-4}$. Since $\alpha^3 = 1$ and $\beta^3 = 1$,we have $\alpha^4 = \alpha^3 \cdot \alpha = \alpha = \omega$ and $\beta^4 = (\omega^2)^4 = \omega^8 = \omega^6 \cdot \omega^2 = \omega^2$. Also,$\alpha^{-4}\beta^{-4} = (\alpha\beta)^{-4} = (\omega \cdot \omega^2)^{-4} = (\omega^3)^{-4} = (1)^{-4} = 1$. Thus,$\alpha^4+\beta^4+\alpha^{-4}\beta^{-4} = \omega + \omega^2 + 1$. Since $1+\omega+\omega^2 = 0$,the value is $0$.

If $\alpha, \beta$ are the roots of $1+x+x^2=0$,then the value of $\alpha^4+\beta^4+\alpha^{-4}\beta^{-4}$ is

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