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. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $1$

Given equation:

$x^2-x+1=0$

Solving it, we get

$x=\frac{1\pm i\sqrt{3}}{2}$

Or, $x=\{w,w^2\}$

$\alpha =-w,\beta =-w^2$

We know that, $w^3=1$, therefore

${\alpha }^{101}+{\beta }^{107}=(-w{)}^{101}+{(-w^2)}^{107}$

$=(-w{)}^{99}\cdot (-w{)}^2+(-w^2{)}^{105}\cdot (-w^2{)}^2$

$=-[w^2+w]=-(-1)$ (sum of roots)

Hence, ${\alpha }^{101}+{\beta }^{107}=1$