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(B) Given equation is $x^{2}+x+1=0$. The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^{2}$. Let $\alpha = \omega$ a

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$-1$

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(B) Given equation is $x^{2}+x+1=0$. The roots of this equation are the complex cube roots of unity,$\omega$ and $\omega^{2}$. Let $\alpha = \omega$ and $\beta = \omega^{2}$. We need to find $\alpha^{16} + \beta^{16} = \omega^{16} + (\omega^{2})^{16}$. Since $\omega^{3} = 1$,we have $\omega^{16} = (\omega^{3})^{5} \cdot \omega = 1^{5} \cdot \omega = \omega$. Similarly,$\omega^{32} = (\omega^{3})^{10} \cdot \omega^{2} = 1^{10} \cdot \omega^{2} = \omega^{2}$. Thus,$\alpha^{16} + \beta^{16} = \omega + \omega^{2}$. Since $1 + \omega + \omega^{2} = 0$,it follows that $\omega + \omega^{2} = -1$.

If $\alpha$ and $\beta$ are the roots of $x^{2}+x+1=0$,then $\alpha^{16}+\beta^{16}$ is equal to

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