Theory of Equations 1 Question 13
14. If $\alpha, \beta \in C$ are the distinct roots of the equation $x^{2}-x+1=0$, then $\alpha^{101}+\beta^{107}$ is equal to
(2018 Main)
(a) -1
(b) 0
(c) 1
(d) 2
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Solution:
- We have, $\alpha, \beta$ are the roots of $x^{2}-x+1=0$
$\because$ Roots of $x^{2}-x+1=0$ are $-\omega,-\omega^{2}$
$$ \begin{aligned} & \therefore \quad \text { Let } \alpha=-\omega \text { and } \beta=-\omega^{2} \\ & \begin{array}{rlr} \Rightarrow \alpha^{101}+\beta^{107} & =(-\omega)^{101}+\left(-\omega^{2}\right)^{107}=-\left(\omega^{101}+\omega^{214}\right) \\ & =-\left(\omega^{2}+\omega\right) & {\left[\because \omega^{3}=1\right]} \\ & =-(-1) & {\left[\because 1+\omega+\omega^{2}=0\right]} \end{array} \\ & =1 \end{aligned} $$
