If , C are the distinct roots of x^2-x+1=0, then ^101+^107 is equal to:

Question

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

Accepted Answer

$\alpha=-\omega,\ \beta=-\omega^2$ (roots of $x^2-x+1=0$).

$$\alpha^{101}+\beta^{107}=(-\omega)^{101}+(-\omega^2)^{107}=-\omega^{101}-\omega^{214}=-\omega^2-\omega=1$$

(using $\omega^{101}=\omega^{2}$, $\omega^{214}=\omega^{1}$, and $1+\omega+\omega^2=0$).

Solutions · 1