Question

If $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$ then ${\alpha }^{16}+{\beta }^{16}$=

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

Answer: (C)

Given, equation is $x^2+x+1=0$
$\Rightarrow x=\frac{-1\pm \sqrt{3}i}{2}$
$\Rightarrow x=\omega ,{\omega }^2$
Since, $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$
$\therefore \alpha =\omega$ and $\beta ={\omega }^2$
Now, ${\alpha }^{16}+{\beta }^{16}=(\omega {)}^{16}+{\left({\omega }^2\right)}^{16}$
$={\omega }^{16}+{\omega }^{32}$
$=\omega +{\omega }^2\left(\because {\omega }^3=1\right)$
$=-1\left(\because 1+\omega +{\omega }^2=0\right)$