Why isn't $x^2+x+1$ a factor of $x^{12}+x^6+1$? When we solve the equation
$$x^{12}+x^6+1=0$$ we obtain $2$ solutions that also satisfy $$x^2+x+1=0$$
namely $-\frac{1}{2}\pm i\frac{\sqrt3}{2}$.
Shouldn't this imply that $x^2+x+1$ is a factor of $x^{12}+x^6+1$? However, the fully factorsied form of
$x^{12}+x^6+1$ is $$(x^6-x^3+1)(x^6+x^3+1)$$
The reason I think that it should be a factor is that when we have, for a function $f(x)$, (here a polynomial,) $f(a)=0$ then we know that $(x-a)$ is a factor of $f(x)$.
If we multiply $(x-(-\frac{1}{2}+ i\frac{\sqrt3}{2}))$ by $(x-(-\frac{1}{2}- i\frac{\sqrt3}{2}))$ which are both factors of $x^{12}+x^6+1$ we get $x^2+x+1$. It seems to me that this should apply to the polynomials above. I have a feeling that my logic is specious, but I'm not entirely certain why; I think the answer to my problem may be that $x^2+x+1$ is factor of $x^{12}+x^6+1$, but only if the other factor has complex coefficients, but I'm not sure. Thank you for your help.
EDIT
Oh wow, I'm so sorry for making such a stupid mistake everyone, thanks for correcting me (see numerous comments and answers). I was looking at a few polynomials simultaneously and got mixed up. Thanks for your help.
 A: You are correct that if all of the roots of a polynomial $f(x)$ are roots of a polynomial $g(x)$, then $f(x)$ is a factor of $g(x)$.
However $x^2 + x + 1$ is not a factor of $x^{12} + x^6 + 1$. The reason is because $- \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$ are not roots of that polynomial. This is because
$$ \left( - \frac{1}{2} \pm i \frac{\sqrt{3}}{2} \right)^3 = 1,$$
so plugging it into $x^{12} + x^6 + 1$ gives $(x^3)^4 + (x^3)^2 + 1 = 1 + 1 + 1 = 3 \neq 0$.
A: I'm sorry, but I don't know how you arrived at your assertion that $-1/2 \pm i \sqrt{3}/2$ are roots of $x^{12} + x^6 + 1$.  In fact, we note $$-\frac{1}{2} \pm i \frac{\sqrt{3}}{2} = e^{\pm 2\pi i/3},$$ hence $$(e^{\pm 2\pi i/3})^{12} + (e^{\pm 2\pi i/3})^6 + 1 = 1 + 1 + 1 = 3 \ne 0.$$
A: @AnginaSeng has already proved dividing $p(x):=x^{12}+x^6+1$ by $x^2+x+1$ leaves a remainder of $3$.
In thinking the remainder would be $0$, you seem to have misindentified the roots of $p$. Since $x^6=\exp\frac{\pm2\pi i}{3}$, $x=\exp\frac{\pm\pi i}{9}\exp\frac{n\pi i}{3}$ for $n\in\{0,\,1,\,2,\,3,\,4,\,5\}$. You can verify, e.g. by working modulo $\frac{\pi i}{9}$, that none of these $12$ roots' phases matches either root of $x^2+x+1$, which would be $\exp\frac{\pm 2\pi i}{3}=\exp\frac{\pm6\pi i}{9}$.
A: Another issue besides that mentioned by several posters in both answers and comments is this: You say the "the fully factorsied form" is a certain product of two cubic polynomials with integer coefficients. But if that is "the fully factorsied form", that would mean that's as much factoring as can be done with integer coefficients. With complex coefficients, a cubic can be factored further.
A: I just want to note a technique based on my comment to the original post where a factor of $x^2+x+1$ can be shown longhand for $x^{2n}+x^n+1$ where $n$ is not a multiple of $3$.
Suppose $n=3m+1$ then $$x^{6m+2}+x^{3m+1}+1=x^{6m+2}-x^2+x^{3m+1}-x+x^2+x+1=$$$$=x^2\left(x^{6m}-1\right)+x\left(x^{3m}-1\right)+x^2+x+1$$ and the two bracketed terms are divisible by $x^3-1=(x-1)(x^2+x+1)$
With $n=3m+2$ we have similarly $$x^{6m+4}+x^{3m+2}+1=x^{6m+4}-x+x^{3m+2}-x^2+x^2+x+1=$$$$=x\left(x^{6m+3}-1\right)+x^2\left(x^{3m}-1\right)+x^2+x+1$$ and a factor $x^3-1$ is there for the two bracketed terms.
The various other methods are more efficient, but this is direct.
