X3-4x-1=0

Step 1: The given equation is the depressed cubic
$x^3 + px + q = 0$
where $p = -4$ and $q = -1$.

Step 2: Compute the discriminant
$D = \left( \frac{q}{2} \right)^2 + \left( \frac{p}{3} \right)^3.$
Substitute the values:
$\frac{q}{2} = -\frac{1}{2}, \quad \left( \frac{q}{2} \right)^2 = \frac{1}{4},$
$\frac{p}{3} = -\frac{4}{3}, \quad \left( \frac{p}{3} \right)^3 = -\frac{64}{27}.$
Then
$D = \frac{1}{4} - \frac{64}{27} = \frac{27}{108} - \frac{256}{108} = -\frac{229}{108} < 0.$
Since $D < 0$, there are three distinct real roots.

Step 3: Use the trigonometric formula for the roots:
$x_k = 2 \sqrt{ -\frac{p}{3} } \cos\left( \frac{1}{3} \arccos\left( \frac{ -\frac{q}{2} }{ \left( -\frac{p}{3} \right)^{3/2} } \right) - \frac{2\pi k}{3} \right), \quad k = 0,1,2.$

Step 4: Compute the amplitude factor:
$-\frac{p}{3} = \frac{4}{3}, \quad \sqrt{ -\frac{p}{3} } = \sqrt{ \frac{4}{3} } = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3},$
$2 \sqrt{ -\frac{p}{3} } = 2 \cdot \frac{2\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}.$

Step 5: Compute the cosine argument:
$-\frac{q}{2} = \frac{1}{2},$
$\left( -\frac{p}{3} \right)^{3/2} = \left( \frac{4}{3} \right)^{3/2} = \frac{4}{3} \cdot \frac{2}{\sqrt{3}} = \frac{8}{3\sqrt{3}}.$
Then
$\frac{ -\frac{q}{2} }{ \left( -\frac{p}{3} \right)^{3/2} } = \frac{1/2}{8/(3\sqrt{3})} = \frac{1}{2} \cdot \frac{3\sqrt{3}}{8} = \frac{3\sqrt{3}}{16}.$
Let $\theta = \arccos\left( \frac{3\sqrt{3}}{16} \right)$.

Step 6: The roots are
$x_k = \frac{4\sqrt{3}}{3} \cos\left( \frac{\theta}{3} - \frac{2\pi k}{3} \right), \quad k=0,1,2.$

The three real roots are
$\boxed{ x_0 = \dfrac{4\sqrt{3}}{3} \cos\left( \dfrac{1}{3} \arccos \dfrac{3\sqrt{3}}{16} \right),\quad x_1 = \dfrac{4\sqrt{3}}{3} \cos\left( \dfrac{1}{3} \arccos \dfrac{3\sqrt{3}}{16} - \dfrac{2\pi}{3} \right),\quad x_2 = \dfrac{4\sqrt{3}}{3} \cos\left( \dfrac{1}{3} \arccos \dfrac{3\sqrt{3}}{16} - \dfrac{4\pi}{3} \right) }$