Let alpha and beta be the roots of x^{2}+x+1= | vedclass

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(A) Given the quadratic equation $x^{2}+x+1=0$.Using the quadratic formula, the roots are $x = \frac{-1 \pm \sqrt{1^{2}-4(1)(1)}}{2} = \frac{-1 \pm \s

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$2 \cos \left(\frac{2 n \pi}{3}\right)$

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(A) Given the quadratic equation $x^{2}+x+1=0$.Using the quadratic formula, the roots are $x = \frac{-1 \pm \sqrt{1^{2}-4(1)(1)}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$.Let $\alpha = \frac{-1+i\sqrt{3}}{2} = e^{i(2\pi/3)}$ and $\beta = \frac{-1-i\sqrt{3}}{2} = e^{-i(2\pi/3)}$.Then, $\alpha^{n}+\beta^{n} = (e^{i(2\pi/3)})^{n} + (e^{-i(2\pi/3)})^{n} = e^{i(2n\pi/3)} + e^{-i(2n\pi/3)}$.Using Euler's formula, $e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$.Therefore, $\alpha^{n}+\beta^{n} = 2\cos\left(\frac{2n\pi}{3}ight)$.

Let $\alpha$ and $\beta$ be the roots of $x^{2}+x+1=0$. If $n$ is a positive integer, then $\alpha^{n}+\beta^{n}$ is

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