Question

Let $\omega \ne 1$ 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

Answer: 33

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