Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here. 


  The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....


I tried the polynomial remainder theorem but I am not sure if I did it correctly.

By factor theorem definition, provided by Wikipedia,


  the remainder of the division of a polynomial $f(x)$ by a linear polynomial  $x-r$ is equal to $f(r)$.


So I attempted to find $r$ by factorizing $x^2-x+1$ first but I got the complex form $x=\frac{1\pm\sqrt{3}i}{2}=r$. 

$f(r)$ is then $(\frac{1+\sqrt{3}i}{2})^3$ or $(\frac{1-\sqrt{3}i}{2})^3$ which do not sound right.

However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help.
Since $$x^3+1 = (x+1)(x^2-x+1)$$ so $$x^3 = (x+1)(x^2-x+1)-1$$ the answer is $-1$.

Similarly for \begin{eqnarray}x^{3n}+1 &amp;=&amp; (x^3+1)\underbrace{\Big((x^3)^{n-1}-(x^3)^{n-2}+...-(x^3)+1\Big)}_{q(x)}\\
&amp;=&amp; (x+1)(x^2-x+1)q(x)\\
\end{eqnarray}

so the answer is again $-1$.
(1) Since $x^3=(x^3+1)-1=(x+1)(x^2-x+1)-1$, the remainder is $-1$.

(2) $\displaystyle x^{2007}=(x^3)^{669}=[(x+1)(x^2-x+1)-1]^{669}=-1+\sum_{k=1}^{669}(x+1)^k(x^2-x+1)^k(-1)^{669-k}$

The remainder is $-1$.