Question

If $\alpha$ and $\beta$ are roots of $x^2-x+1=0$, then the value of ${\alpha }^{2013}+{\beta }^{2013}$ is

• A. $2$
• B. $-2$
• C. $-1$
• D. $1$

Answer 1

The correct option is B $-2$

Given equation is $x^2-x+1=0$

$x=\frac{1\pm \sqrt{1-4}}{2}(\because x=\frac{b\pm \sqrt{b^2-4ac}}{2a})$

$=\frac{1\pm i\sqrt{3}}{2}=\frac{1+i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}$

$\Rightarrow -x=\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$$\Rightarrow +x=\omega ,{-\omega }^2$

Since $(\alpha ,\beta$ are the roots of given equation)

Then, $\alpha =-\omega$ and $\beta =-{\omega }^2$

$\therefore {\alpha }^{2013}+{\beta }^{2013}=-{\left(\omega \right)}^{2013}+{(-\omega )}^{2013}$

$={-\omega }^{2013}-{\omega }^{4026}$

$={-\left({\omega }^3\right)}^{671}{-\left({\omega }^3\right)}^{1342}$

$={-\left(1\right)}^{671}-{\left(1\right)}^{1342}(\because {\omega }^3=1)$

$=-1-1=-2$