If alpha is a root of the equation x^2-x+1=0, | vedclass

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(C) Given the equation $x^2-x+1=0$. The roots are $\alpha = -\omega$ and $\alpha = -\omega^2$,where $\omega$ is the complex cube root of unity. Since

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(C) Given the equation $x^2-x+1=0$. The roots are $\alpha = -\omega$ and $\alpha = -\omega^2$,where $\omega$ is the complex cube root of unity. Since $\alpha^2-\alpha+1=0$,we have $\alpha+\frac{1}{\alpha} = 1$. Let $S_n = \alpha^n + \frac{1}{\alpha^n}$. For $n=1$,$S_1 = \alpha + \frac{1}{\alpha} = 1$. For $n=2$,$S_2 = \alpha^2 + \frac{1}{\alpha^2} = (\alpha+\frac{1}{\alpha})^2 - 2 = 1^2 - 2 = -1$. For $n=3$,$S_3 = \alpha^3 + \frac{1}{\alpha^3} = (-1)^3 + \frac{1}{(-1)^3} = -1 - 1 = -2$. For $n=4$,$S_4 = \alpha^4 + \frac{1}{\alpha^4} = \alpha + \frac{1}{\alpha} = 1$. For $n=5$,$S_5 = \alpha^5 + \frac{1}{\alpha^5} = \alpha^2 + \frac{1}{\alpha^2} = -1$. For $n=6$,$S_6 = \alpha^6 + \frac{1}{\alpha^6} = 1 + 1 = 2$. The terms are $S_n^3$. $S_1^3 = 1^3 = 1$,$S_2^3 = (-1)^3 = -1$,$S_3^3 = (-2)^3 = -8$,$S_4^3 = 1^3 = 1$,$S_5^3 = (-1)^3 = -1$,$S_6^3 = 2^3 = 8$. The sequence of terms is $1, -1, -8, 1, -1, 8, 1, -1, -8, 1, -1, 8$. Summing these $12$ terms: $(1-1-8) + (1-1+8) + (1-1-8) + (1-1+8) = -8 + 8 - 8 + 8 = 0$.

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3+\ldots$ to $12$ terms $=$

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