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.

  • $\begingroup$ This question is always changing and none has a clear meaning. $\endgroup$ – Shuchang Dec 28 '13 at 10:26
  • $\begingroup$ Actually i am trying to get x raised to power 2013 $\endgroup$ – kartikeykant18 Dec 28 '13 at 10:27
  • $\begingroup$ OK, I get it. Try x^{2013}. $\endgroup$ – Shuchang Dec 28 '13 at 10:29
  • $\begingroup$ Related : math.stackexchange.com/questions/585425/… $\endgroup$ – lab bhattacharjee Dec 28 '13 at 10:57
  • $\begingroup$ 2013's gone, un‐mathematically speaking, maybe 2014 would be more fashionable. Oh well, the answer is 2014! Cool. $\endgroup$ – Sawarnik Dec 28 '13 at 11:43


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


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 $

  • 2
    $\begingroup$ Welcome to MSE, Please refrain from using greetings in your answers. Answers are supposed to be timeless and formal. I speak on behalf of the entire community when I say, we appreciate your enthusiasm and would love to see more from you. Happy New Year! (comments are different... sorry for being a hypocrite) $\endgroup$ – Nick Dec 28 '13 at 11:38

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 $\,x^2+x+1\mid\color{#c00}{ x^3-1}\mid \color{#0a0}{x^{3k}-1}$


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