First, using the usual way by assuming $x^3-3x+1=(x+a)(x^2+bx+c)$ and trying to determinne $a$, $b$, and $c$ will lead to solving the original equation $x^3-3x+1=0$, i.e. entering a loop.

In order to solve this problem, let's first study the function $f(x) = x^3-3x+1$. As $f'(x) = 3x^2 -3$, we find $f(x)$ has to local extreme values at $x=\pm 1$. Furthermore, we found $f(x) is decreasing when $x< -1$, and increasing when $-1 < x < 1$, and decreasing again when $x > 1$.

Meanwhile, we find that $f(-2) < 0$, $f(-1) > 0$, $f(1) < 0 $, and $f(2) > 0$. Therefore, we conclude that all its roots lie between $-2$ and $2$. As such, let's assume $x=2\sin\theta$ where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. It follows that: $$x^3-3x+1 = 8\sin^3\theta - 6\sin\theta + 1= 2(4\sin^3\theta - 3\sin\theta)+1 = -2\sin3\theta + 1= 0$$

Therefore, $$\sin3\theta=\frac{1}{2}\implies 3\theta = \frac{\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}\implies \theta=\frac{\pi}{18}, \frac{5\pi}{18}, -\frac{7\pi}{18}$$

Finally, the three roots to the original equations are $2\sin\frac{\pi}{18}$, $2\sin\frac{5\pi}{18}$, and $-2\sin\frac{7\pi}{18}$, respectively.