X3-6x+3=0

Step 1: The given equation is the depressed cubic $x^3 - 6x + 3 = 0$ or $x^3 + px + q = 0$ where $p = -6$, $q = 3$.

Step 2: Compute the discriminant $\Delta_0 = \left( \frac{q}{2} \right)^2 + \left( \frac{p}{3} \right)^3.$ Substitute $q = 3$, $p = -6$: $\frac{q}{2} = \frac{3}{2}, \quad \left( \frac{3}{2} \right)^2 = \frac{9}{4},$ $\frac{p}{3} = \frac{-6}{3} = -2, \quad (-2)^3 = -8.$ Then $\Delta_0 = \frac{9}{4} - 8 = \frac{9}{4} - \frac{32}{4} = -\frac{23}{4} < 0.$ Since $\Delta_0 < 0$, there are three distinct real roots.

Step 3: Compute $r = \sqrt{ -\frac{p}{3} }.$ Substitute $p = -6$: $-\frac{p}{3} = \frac{6}{3} = 2, \quad r = \sqrt{2}.$

Step 4: Compute $\cos \phi = \frac{ -q/2 }{ r^3 }.$ First, $r^3 = (\sqrt{2})^3 = 2\sqrt{2},$ $-\frac{q}{2} = -\frac{3}{2}.$ Then $\cos \phi = \frac{ -3/2 }{ 2\sqrt{2} } = -\frac{3}{4\sqrt{2}}.$ Rationalize: $-\frac{3}{4\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = -\frac{3\sqrt{2}}{8}.$ So $\phi = \arccos\left( -\frac{3\sqrt{2}}{8} \right).$

Step 5: The roots are $x_k = 2\sqrt{2} \cos\left( \frac{1}{3} \arccos\left( -\frac{3\sqrt{2}}{8} \right) - \frac{2\pi k}{3} \right), \quad k = 0,1,2.$

Final answers: $\boxed{ x = 2\sqrt{2} \cos\left( \frac{1}{3} \arccos\left( -\frac{3\sqrt{2}}{8} \right) - \frac{2\pi k}{3} \right),\ k=0,1,2 }$