6. Factorize $x^{2 m+1}-1\left(m \in \mathbf{N}^{*}\right)$ over the real numbers.
6. Let $1, \varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{2 m}$ be all the $2 m+1$-th roots of unity, where $\varepsilon_{t}=\cos \frac{2 k \pi}{2 m+1}+\mathrm{i} \sin \frac{2 k \pi}{2 m+1}, k=1,2, \cdots, 2 m$. It is given that $\varepsilon_{1} \varepsilon_{2 m+1}=1(k=1,2, \cdots, 2 m)$. Also, $\varepsilon_{r}+\varepsilon_{2 m,-1}=\varepsilon_{c}+\overline{\varepsilon_{t}}=2 \cos \frac{2 k \pi}{2 m+1}$. Therefore,
$$
\begin{aligned}
x^{2 m+1}-1 & =(x-1)\left(x-\varepsilon_{1}\right)\left(x-\varepsilon_{2}\right) \cdots\left(x-\varepsilon_{2 m-1}\right)\left(x-\varepsilon_{2 m}\right) \\
& =(x-1)\left[\left(x-\varepsilon_{1}\right)\left(x-\varepsilon_{2 m}\right)\right]\left[\left(x-\varepsilon_{2}\right)\left(x-\varepsilon_{2 m-1}\right)\right] \cdots\left[\left(x-\varepsilon_{m}\right)\left(x-\varepsilon_{m+1}\right)\right] \\
& =(x-1)\left[x^{2}-2\left(\cos \frac{2 \pi}{2 m+1}\right) x+1\right] \cdots\left[x^{2}-2\left(\cos \frac{2 m \pi}{2 m+1}\right) x+1\right] .
\end{aligned}
$$