Question

If $\alpha$ is a non-real root of $x^6=1$, then $\frac{{\alpha }^5+{\alpha }^3+\alpha +1}{{\alpha }^2+1}=$

• A. ${\alpha }^2$
• B. $0$
• C. $-{\alpha }^2$
• D. $\alpha$

Answer 1

Answer: (C)

$-{\alpha }^2$

$1+\alpha +...+{\alpha }^5=0$ [sum of n roots of unity]
$\Rightarrow 1+\alpha +{\alpha }^3+{\alpha }^5=-({\alpha }^2+{\alpha }^4)$
$\Rightarrow 1+\alpha +{\alpha }^3+{\alpha }^5=-{\alpha }^2\left({1+\alpha }^2\right)$
$\Rightarrow \frac{1+\alpha +{\alpha }^3+{\alpha }^5}{{1+\alpha }^2}=-{\alpha }^2$