Where I made a mistake during factoring $x^6+x^5+x^4+x^3+x^2+x+1$? In order to factor the expression, due to symmetry of coefficients if $r_1,r_2,r_3$ are zeros of  $x^6+x^5+x^4+x^3+x^2+x+1$ then $\frac{1}{r_1}, \frac{1}{r_2} , \frac{1}{r_3}$ are also zeros. So we can rewrite:
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2-(r_1+\frac{1}{r_1})x+1)+(x^2-(r_2+\frac{1}{r_2})x+1)(x^2-(r_3+\frac{1}{r_3})x+1)$$
And it can be written as:$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)(x^2+cx+1)$$
$$=x^6+(a+b+c)x^5+(3+ab+ac+bc)x^4+(abc+a+b+c)x^3+(3+ab+ac+bc)x^2+(a+b+c)x+1$$
In orther to find$a,b,c$ I should solve the system :
${ \begin{cases}{a+b+c=1} \\ {ab+ac+bc=-2} \\ {abc=0}\end{cases} }$
from the equation $abc=0$ We have $a=0$ (Since our derivation was symmetric in $a,b,c$ it doesn't matter to put $a=0$ or $b=0$ or $c=0$).
${ \begin{cases}{b+c=1} \\ {bc=-2}\end{cases} }$
and from that we have $b=2$ , $c=-1$ , $a=0$
So I get :
$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+1)(x^2+2x+1)(x^2-x+1)=(x^2+1)(x+1)^2(x^2-x+1)$$
But my answer is wrong because if we multiply the equation by $x-1$ We should obtain $x^7-1$. But:
$$(x-1)(x^2+1)(x+1)^2(x^2-x+1)=(x-1)(x+1)(x^2+1)(x+1)(x^2-x+1)=(x^4-1)(x^3+1) \ne x^7-1$$
I checked my answer several times but I couldn't find my mistake.

To factor over $\mathbb{R}$, note that:
$$
\begin{align}
x^6+\cdots+1
&=\frac{1}{x-1}\left(x^7-1\right)
\end{align}
$$
where $x^7-1$ has its roots being the complex seventh roots of unity. One such root is of course $1$, while the other six pair up with their complex conjugate:
$$
\begin{align}
&x^6+\cdots+1\\
&=\frac{1}{x-1}(x-1)\left(x-e^{2\pi i/7}\right)\left(x-e^{-2\pi i/7}\right)\left(x-e^{4\pi i/7}\right)\left(x-e^{-4\pi i/7}\right)\left(x-e^{6\pi i/7}\right)\left(x-e^{-6\pi i/7}\right)\\
&=\left(x^2-2\cos(2\pi/7)x+1\right)\left(x^2-2\cos(4\pi/7)x+1\right)\left(x^2-2\cos(6\pi/7)x+1\right)
\end{align}
$$
This is the full factorization over $\mathbb{R}$.
Any factorization over $\mathbb{Q}$ would have to be composed of polynomials where complex conjugate roots are paired. So a factorization over $\mathbb{Q}$ is either trivial (the original polynomial is irreducible) or with all other factorings under consideration, at least one of those quadratic polynomials would be present, and  one of $\cos(2\pi/7)$, $\cos(4\pi/7)$, $\cos(6\pi/7)$ is rational. But the only way that the cosine of a rational multiple of $\pi$ can be a rational number is if said cosine is $0$, $\pm\frac12$, or $\pm1$. (This last fact not trivial, just something I've memorized. A proof is accessible though.) And of course those three cosine values are not in $\{0,\pm\frac12,\pm1\}$. Therefore the original polynomial must be irreducible.
