Finding $n$ such that $(3-3w+2w^2)^{4n+3}+(2+3w-3w^2)^{4n+3}+(-3+2w+3w^2)^{4n+3}=0$ for $w\neq1$ a cube root of unity 
Let
$${\left( {3 - 3\omega  + 2{\omega ^2}} \right)^{4n + 3}} + {\left( {2 + 3\omega  - 3{\omega ^2}} \right)^{4n + 3}} + {\left( { - 3 + 2\omega  + 3{\omega ^2}} \right)^{4n + 3}}=0$$
If $\omega\ne1$ be the complex cube root of unity, then the possible values of $n$ is
(A) $1\quad$ (B) $2\quad$ (C) $3\quad$ (D) 4

My approach $1+\omega+\omega^2=0$
Let $A={\left( {3 - 3\omega  + 2{\omega ^2}} \right)}$
$B={\left( {2 + 3\omega  - 3{\omega ^2}} \right)} $
$C= {\left( { - 3 + 2\omega  + 3{\omega ^2}} \right)}$
$B=A\omega$, $C=A\omega^2$ as $A+B+C=0$. Not able to proceed

Your sum is $A^{4n+3}(1+\omega^{4n+3}+\omega^{2(4n+3)})$, and $A\ne 0$. So we must have $1+\omega^{4n+3}+\omega^{2(4n+3)}=0$. And $\omega^3=1$. So this is
$$1+\omega^n+\omega^{2n}=0$$
which is true if and only if $n\equiv 1$ or $2\bmod 3$.
As to which of the four answers to select, that is a different puzzle...
