Question

If $\alpha$ and $\beta \in C$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. $-2$
• D. $0$

Answer 1

The correct option is A

$1$

Explanation for correct option:

Step1. Find the root of the equation :

Given that $\alpha$ and $\beta \in C$are the distinct roots of the equation$x^2-x+1=0$.

$$\begin{gathered} x^2-x+1=0 \\ \Rightarrow x=\frac{1\pm \sqrt{1-4}}{2} \\ \Rightarrow x=\frac{1\pm i\sqrt{3}}{2} \\ \Rightarrow x=\frac{1+i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2} \\ \Rightarrow x=-{\omega }^2,-\omega \end{gathered}$$

$\Rightarrow \alpha =-{\omega }^2;\beta =-\omega$

Step2. Find the value of ${\mathit{\alpha }}^{\mathbf{101}}\mathbf{+}{\mathit{\beta }}^{\mathbf{107}}$:

$$\begin{gathered} {\alpha }^{101}+{\beta }^{107}={(-{\omega }^2)}^{101}+{(-\omega )}^{107}\Rightarrow {\alpha }^{101}+{\beta }^{107}=-({\omega }^{202}+{\omega }^{107}) \\ \Rightarrow {\alpha }^{101}+{\beta }^{107}=-[{\left({\omega }^3\right)}^{67}\omega +{\left({\omega }^3\right)}^{35}{\omega }^2] \\ \Rightarrow {\alpha }^{101}+{\beta }^{107}=-[\omega +{\omega }^2] \\ \Rightarrow {\alpha }^{101}+{\beta }^{107}=1 \end{gathered}$$

Hence, the correct option is A.