<s>[INST] Why is the "solving for cubic equation roots general rule" sometimes not applicable while the equation obviously has roots? the general rule:
we have $ax^3+bx^2+cx+d=0$
$\Delta_0=b^2-3ac$
$\Delta_1=2b^3-9abc+27a^2d$
$C=\sqrt{\Delta_1^2-4\Delta_0^3}$
$D=(\frac{\Delta_1+C}{2})^\frac{1}{3}$
$x=-\frac{1}{3a}(b+D+\frac{\Delta_0}{D})$
imagine $x^3-6x^2+11x-6=0$
we know its roots are $x=1$, $x=2$ and $x=3$.
but when you use the general rule, you will find a negative $\Delta_1^2-4\Delta_0^3$ and thus; you can't continue the process!
Also, when you use this rule, you will just find ONE real root (remember the last process to find final $x$); while that equation has 3 real roots and no imaginary roots. so how to find the other real roots using the general rule?!
 [/INST] 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: $C = i\sqrt{27}$, $D = \eta \sqrt{3}/\sqrt[3]{2}$. 
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