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.
 A: 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 &=& (x^3+1)\underbrace{\Big((x^3)^{n-1}-(x^3)^{n-2}+...-(x^3)+1\Big)}_{q(x)}\\
&=& (x+1)(x^2-x+1)q(x)\\
\end{eqnarray}
so the answer is again $-1$.
A: Another way if you're familiar with modular arithmetic is to work modulo $x^2-x+1$, in which case we have $x^2\equiv x-1$ and thus
$$x^3\equiv xx^2\equiv x(x-1)\equiv x^2-x\equiv (x-1)-x\equiv -1.$$ This can be extended to your other question by noting that $x^{2007}=\left(x^3\right)^{669}$.
A: One way of writing this is to borrow the notion of equivalence (encoded in the notion of Ideals etc)
Because $x^3+1=(x+1)(x^2-x+1)$, we can write $x^3\equiv -1 \bmod (x^2-x+1)$
Then $x^{2007}=(x^3)^{669}\equiv (-1)^{669}$
This can be a surprisingly effective and efficient way of doing these questions about polynomial division and remainders.
A: (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 $-1$.
