Find $x ^{ 2013} + 2013x ^{ 2010}$

Question

Q. If $\large {\space x^2 + x + 1 = 0\space } $, Find $ x ^{ 2013} + 2013x ^{ 2010}$.

I have tried finding the roots of $x$ from the given equation but that does not work.

Answer 1

HINT:

As $\displaystyle x^2+x+1=0, x^3-1=(x-1)(x^2+x+1)=0\implies x^3=1$

$\displaystyle \implies x^{3m}=(x^3)^m=1$ if $m$ is an integer

Answer 2

Let $\large\omega$ and $\large\omega^2$ be the roots of $ x^2+x+1$.

$\large\omega^3=1$ and $(\large\omega^2)^3=1$.

As 2013 is divisible by 3,
$\space \large\omega^{2013}=1$.

Similarly, $\large\omega^{2010}=1$.

For the case $\large\omega^2$ we argue similarly.
Thus, the answer is $ 1+ 2013 =2014 $

Answer 3

If modular arithmetic is unfamiliar then you may poceed as follows

$$x^{3n} + 3n\, x^{3n-3}\, =\, x^{3n-3} (\color{#c00}{x^3-1}) + (3n+1) (\color{#0a0}{x^{3(n-1)}-1}) + 3n+1$$

Notice the colored terms are $= 0\,$ since $\,x^2+x+1\mid\color{#c00}{ x^3-1}\mid \color{#0a0}{x^{3k}-1}$