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} $
$=\omega^{16}+\omega^{32}$
$=\omega+\omega^{2} \,\,\,\,\left(\because \omega^{3}=1\right)$
$=-1 \,\,\,\,\,\,\, \left(\because 1+\omega+\omega^{2}=0\right)$