Question

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

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

Answer 1

The correct option is B $-{\alpha }^2$

$\because x^6=1$
For the given equation we know that sum of roots is equal to zero
$\therefore 1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5=0$
$\Rightarrow {\alpha }^5+{\alpha }^3+\alpha +1=-{\alpha }^4-{\alpha }^2\cdots \left(1\right)$
Now,
$\frac{{\alpha }^5+{\alpha }^3+\alpha +1}{{\alpha }^2+1}$
$=\frac{-{\alpha }^4-{\alpha }^2}{{\alpha }^2+1}\text{(using (1))}$
$=\frac{-{\alpha }^2({\alpha }^2+1)}{{\alpha }^2+1}$
$=-{\alpha }^2$