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

Question

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

Would it be ended up in this form $A+B+C=A^{4n+3}(1+\omega^{4n}+\omega^{8n})=0$? — MathArt, Dec 14 '20 at 13:54
(i) Using $A,B,C$ as variable names when you already have them as answer options is liable to lead to confusion. (ii) I think you have to translate the question better: "the possible values of $n$ is" makes no sense. — TonyK, Dec 14 '20 at 14:07

score 3 · Accepted Answer · answered Dec 14 '20 at 14:09

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...

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

1 Answers1