When applying Cardano's formulae, you must take specific branches of the square and cubic root functions. Also, note that the square root of 108 is justs a single value; while $x^2 - 108=0$ has two (real) solutions, the expression $\sqrt{108}$ is a single value: the nonnegative solution to $x^2-108=0$.
In any case, remember that if you have a depressed cubic, $x^3 + px + q = 0$, we introduce two variables $u$ and $v$ subject to the condition that $u+v = x$ and $3uv + p = 0$. This requires the product of $u$ and $v$ to be a real number. Eventually, you arrive at $u$ and $v$ being the cubic roots in question.
But remember that you are assuming that $uv$ is a real number. This is why you cannot just take all possible combinations with all possible roots. In other words, all those "other values" of $u+v$ are values that do not satisfy the system of equations you have:
\begin{align*}
(u+v)^3 + p(u+v) + q &=0\\
3uv + p &=0
\end{align*}
There is no general notation that tells you which branches of the complex-valued cubic and square root functions you must take, you just have to remember that the product of the two cubic roots has to be a real number. This is what cuts down from nine to three values.
There is a standard way of listing all solutions: once you find one value of $u$ and $v$ such that $uv$ is a real number, then the three solutions are given by $u+v$, $\omega u + \omega^2 v$, and $\omega^2u + \omega v$, where $\omega = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$ is a primitive cubic root of unity. (Notice that if $u$ is one value of the cubic root, then the other two values are $\omega u$ and $\omega^2u$, and likewise with $v$).
It seems that you are not quite aware of all the assumptions you are making about $u$ and $v$ when you use Cardano's formula. For one thing, if you review the derivation of the formula, say, in
Wikipedia, you will see that the condition that $3uv+p=0$ is worked into the derivation. Here's the method: remember that $x=u+v$ is supposed to be a solution of
$$x^3 + px + q = 0.$$
Plugging in and doing a bit of work, we get:
\begin{align*}
0 &= (u+v)^3 + p(u+v) + q\\
&= u^3 + 3u^2v + 3uv^2 + v^3 + p(u + v) + q\\
&= u^3 + v^3 + 3uv(u+v) + p(u+v) + q\\
&= u^3 + v^3 + (3uv+p)(u+v) + q.
\end{align*}
It is at this point that the extra condition $3uv+p=0$ is introduced, so that the equation reduces to $u^3+v^3 + q = 0$, Then $u^3+v^3 = -q$, and $(uv)^3 = -\frac{p^3}{27}$. Now remember that if $a$ and $b$ are two numbers, then $(z-a)(z-b) = z^2 - (a+b)z + ab$, so $a$ and $b$ are always the roots of the quadratic that has linear coefficient minus their sum, and constant coefficient their product. Applying this to $u^3$ and $v^3$, since we know their sum and their product, we deduce that $u^3$ and $v^3$ are the two roots of the quadratic:
$$z^2 +qz - \frac{p^3}{27}=0$$
and from this, using the quadratic formula, we obtain that
$$u^3 = -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}, \qquad v^3 = -\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}.$$
Therefore,
$$u = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},\qquad v = \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},$$
so that the solution $x$ is given by the formula you have:
$$x = u+v = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$
But the assumption that $3uv+p=0$ was instrumental in obtaining this formula. The very fact that you write that $x$ is equal to the sum of the two cubic roots
already assumes that the product of those two (complex-valued) cubic roots will be a real number, specifically, $-\frac{p}{3}$.
While you may perhaps not have made that assumption consciously, Cardano certainly did, and by using his formula you are also making the assumption, whether you were aware of it or not.