If alpha, beta in mathbb{C} are distinct root | vedclass

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(B) The roots of the equation $x^2 - x + 1 = 0$ are given by $x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$.These are $-\omega$ and $-

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$1$

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(B) The roots of the equation $x^2 - x + 1 = 0$ are given by $x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$.These are $-\omega$ and $-\omega^2$,where $\omega$ is a complex cube root of unity.Let $\alpha = -\omega$ and $\beta = -\omega^2$.Then $\alpha^{101} + \beta^{107} = (-\omega)^{101} + (-\omega^2)^{107}$.$= -\omega^{101} - \omega^{214}$.Since $\omega^3 = 1$,we have $\omega^{101} = \omega^{3 	imes 33 + 2} = \omega^2$ and $\omega^{214} = \omega^{3 	imes 71 + 1} = \omega$.Thus,$\alpha^{101} + \beta^{107} = -(\omega^2 + \omega)$.Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.Therefore,$\alpha^{101} + \beta^{107} = -(-1) = 1$.

If $\alpha, \beta \in \mathbb{C}$ are distinct roots of the equation $x^2 - x + 1 = 0$,then $\alpha^{101} + \beta^{107}$ is equal to:

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