Question

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

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (C)

$x^2-x+1=0$
Roots are $-\omega ,-{\omega }^2$
Let $\alpha =-\omega ,\beta =-{\omega }^2$
${\alpha }^{101}+{\beta }^{107}=(-\omega {)}^{101}+{\left(-{\omega }^2\right)}^{107}$
$=-\left({\omega }^{101}+{\omega }^{214}\right)$
$=-\left({\omega }^2+\omega \right)$
$=1$