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.

Noe that
$$
(x^2+1)(x+1)^2(x^2-x+1)=
x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1,
$$
so you forgot a factor $2$, as remarked above. The polynomial
$$
\Phi_7(x)=1+x+x^2+x^3+x^4+x^5+x^6
$$
is irreducible over the integers and over the rational numbers by the Gauss lemma and Eisenstein's criterion. Therefore such a factorization cannot hold.
