Why is the "solving for cubic equation roots general rule" sometimes not applicable while the equation obviously has roots?

Yuriy's comment is right. The solutions exist, and the results you get with those complex radicals are such that all the imaginary parts cancel and, as if by magic, they equal the real roots you knew were there all along.

There are two ways to get around the casus irreducibilis. One way is to cheat and find a rational root (if one exists), allowing you to factor the cubic and get the remaining roots from the quadratic quotient. A variation on that is in some applications, the cubic equation is constructed in a way where you know one root, and you can factor with that root. The second approach is to introduce non-algebraic functions, which in this case means trigonometric functions, as described first by Vieta. Details are in the link given above.

You use the Cardano formula to solve a cubic equation. Your formula for $x$ contains real numbers $C$ and $D$ if $\Delta_1^2-4\Delta_0^3 \ge 0$ and thus obviously gives you a real root. If $\Delta_1^2-4\Delta_0^3 > 0$, there are two additional non-real complex roots which are complex conjugate. If $\Delta_1^2-4\Delta_0^3 = 0$, there is one additional real root of multiplicity two.

The case $\Delta_1^2-4\Delta_0^3 < 0$ is known as the casus irreducibilis. In that case there exist three distinct real roots, but your formula represents them via non-real complex numbers $C$ and $D$. See my answer to Is there really analytic solution to cubic equation? where the complete story is told.

Edited:

In your example $x^3- 6x^2 + 11x - 6 = 0$ we have the three real roots $1,2,3$. Transformation as in the above link with $x = y - \frac{1}{3}(-6) = y + 2$ yields $y^3 - y = 0$. This shows that $y = 0$ is a solution and reduces the problem to $y^2 - 1 = 0$ which gives $y = \pm 1$.

Nevertheless we are in the casus irreducibilis: We have $a = -1, b = 0$, thus $R = -1/27 < 0$. Thus $\sqrt{R} = i\sqrt{1/27}$ and $w = \eta /\sqrt{3}$, where $\eta$ is one of the three complex third roots of $i$. We may take $\eta = \sqrt{3}/2 + i/2$ which gives $w = 1/2 + i/2\sqrt{3}$. Thus $w' = 1/(3w) = 1/2 - i/2\sqrt{3}$ and $y = w + w'= 1$ is a solution found by Cardano's formula.

Your formula yields $\Delta_0 = 3, \Delta_1 = 0$. Similar computations as above give a solution of your equation: $C = i\sqrt{27}$, $D = \eta \sqrt{3}/\sqrt[3]{2}$.