Is there a better method to factoring large polynomials such as $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$? The question is as such:

Express $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ in irreducible factors $\in \mathbb{R}$

My initial thoughts to this were to use grouping of factors but this method was to no avail. Instead, I took this detour:
$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ is a factor of $x^7 - 1$, that is $x^7 - 1 = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$
Hence, we can factor $x^7 - 1$ instead and just eliminate $x = 1$.
$$\to x^7 - 1 = 0$$
$$\to x^7 = 1 = e^{i2\pi}$$
$$\to x = e^{i\frac{2k\pi}{7}}, k = \{0, 1, 2, 3, 4, 5, 6\}$$
$$\to x^7 - 1 = (x - 1)(x - e^{i\frac{2\pi}{7}})(x - e^{i\frac{4\pi}{7}})(x - e^{i\frac{6\pi}{7}})(x - e^{i\frac{8\pi}{7}})(x - e^{i\frac{10\pi}{7}})(x - e^{i\frac{12\pi}{7}})$$
Finally we have, $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x - e^{i\frac{2\pi}{7}})(x - e^{i\frac{4\pi}{7}})(x - e^{i\frac{6\pi}{7}})(x - e^{i\frac{8\pi}{7}})(x - e^{i\frac{10\pi}{7}})(x - e^{i\frac{12\pi}{7}})$.
After a bit more thought, we can obtain real factors by group two factors such that upon multiplication, the angle addition obtains $e^{\frac{14\pi}{7}} = 1$, as the $c$ in $x^2 + bx + c$. This would be like $(x - e^{i\frac{2\pi}{7}})(x - e^{i\frac{12\pi}{7}}) = x^2 - 2\sin(\frac{3\pi}{14})x + 1$ (Using $e^{i\theta} = \cos{\theta} + i\sin{\theta}$).
Of course, all of this is tedious by hand to work out and type here, hence my final output was $$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x^2 - 2\sin(\frac{3\pi}{14})x + 1)(x^2 + 2\cos(\frac{\pi}{7})x + 1)(x^2 + 2\sin(\frac{\pi}{14}) + 1)$$.
My Question:
Is this factoring correct, and more importantly is there a more direct/simpler way to factor the polynomial?

Another well-known method is to rewrite the polynomial as follows.
$\begin{align}
&1+z+z^2+z^3+z^4+z^5+z^6=\\
&=z^3\left(\dfrac{1}{z^3}+\dfrac{1}{z^2}+\dfrac{1}{z}+1+z+z^2+z^3\right)=\\
&=z^3\left[\left(\dfrac{1}{z^3}+z^3\right)+\left(\dfrac{1}{z^2}+z^2\right)+\left(\dfrac{1}{z}+z\right)+1\right]=\\
&=z^3\left[\left(\dfrac{1}{z}+z\right)^3-3\left(\dfrac{1}{z}+z\right)+\left(\dfrac{1}{z}+z\right)^2-2+\\
+\left(\dfrac{1}{z}+z\right)+1\right]=\\
&=z^3\left[\left(\dfrac{1}{z}+z\right)^3+\left(\dfrac{1}{z}+z\right)^2-2\left(\dfrac{1}{z}+z\right)-1\right].
\end{align}$
By letting $\;y=\dfrac{1}{z}+z\;$ we get that
$\begin{align}
&1+z+z^2+z^3+z^4+z^5+z^6=z^3\left(y^3+y^2-2y-1\right).
\end{align}$
Now consider the real roots of the polynomial $\;y^3+y^2-2y-1\;$, which all belong to the interval $\;\left]-2,2\right[\; .$
