Question

Let $\omega \neq 1$ be a complex cube root of unity. If $\left(4+5 \omega+6 \omega^{2}\right)^{n^{2}+2}+\left(6+5 \omega^{2}+4 \omega\right)^{n^{2}+2}+\left(5+6 \omega+4 \omega^{2}\right)^{n^{2}+2}=0$, and $n \in N$; where $n \in[1,100]$, then number of values of $n$ is ____

Answer 1

$\left(4+5 \omega+6 \omega^{2}\right)^{n^{2}+2}+\left(6+5 \omega^{2}+4 \omega\right)^{n^{2}+2}+\left(5+6 \omega+4 \omega^{2}\right)^{n^{2}+2} $
$=0 $
$\Rightarrow \left(4+5 \omega+6 \omega^{2}\right)^{n^{2}+2}\left[1+\omega^{n^{2}+2}+\omega^{2 n^{2}+4}\right]=0$
$\Rightarrow n=3 \lambda, n \neq 3 \lambda+1,3 \lambda+2$
$\therefore n=3,6,9,12, \ldots, 99$
So, number of values of $n$ is $33$ .