Question

If $\alpha$ is a non-real root of $x^5+1=0$ then ${\alpha }^{10n+2}+{\alpha }^{5n+2}+{\alpha }^{5n}$, where n is an odd positive integer,has the value

• A. $1$
• B. $0$
• C. $-1$
• D. none of these

Answer 1

The correct option is C $-1$

$x^5=-1$
Hence
${\alpha }^5=-1$
Therefore
${\alpha }^{10n+2}+{\alpha }^{5n+2}+{\alpha }^{5n}$
$=({\alpha }^{5n}{)}^2{\alpha }^2+({\alpha }^{5n}).{\alpha }^2+{\alpha }^{5n}$
$=(-1{)}^2{\alpha }^2+(-1){\alpha }^2+{\alpha }^{5n}$ .... Since n is odd
$=-{\alpha }^2+{\alpha }^2-1$
$=-1$