Let us consider the cubic equation $(1)$ and try to find cubic roots of $-\frac{q}{2} \pm \sqrt{D}$ provided we are given a root $n$ of $(1)$.
Following Thomas Andrews' answer to Denesting Cardano's Formula, we have
$$x^3+px+q=(x-n)(x^2+nx+b)$$
where $b=-\frac{q}{n} = p + n^2$. Note that the quadratic equation $x^2+nx+b = 0$ has the discriminant $-\Delta$, where $\Delta = 4b - n^2 = 4p + 3n^2$. This gives
$$D=\frac{q^2}{4}+\frac{p^3}{27}
= \frac{n^2b^2}{4} +\frac{(b-n^2)^3}{27} = \frac{(4b-n^2)(b+2n^2)^2}{108} \tag{4} \\
= \left(\frac{b+2n^2}{6}\right)^2 \frac{\Delta}{3}.$$
We therefore get
$$\sqrt{D}=\epsilon\frac{b+2n^2}{6}\sqrt{\frac{\Delta}{3}} = \epsilon\frac{p+3n^2}{6}\sqrt{\frac{4p + 3n^2}{3}} \tag{5}$$
with $\epsilon = \operatorname{sign}(p + 3n^2) \in \{-1,0,+1\}$. This is trivial for $D = 0$ since then by $(4)$ any $\epsilon$ will do in $(5)$. For $D \ne 0$, $D$ and $\Delta$ have the same sign by $(4)$, thus both $\sqrt{D}$ and $\sqrt{\frac{\Delta}{3}}$ are either positive real or purely imaginary and our square root convention shows that $\epsilon\frac{b+2n^2}{6}$ must be positive, thus $\epsilon = \operatorname{sign}(b+2n^2) = \operatorname{sign}(p + 3n^2)$.
It is now easy to verify that
$$-\frac{q}{2}\pm\sqrt{D}=\left(\frac{n}{2}\pm \frac{\epsilon}{2}\sqrt{\frac{\Delta}{3}}\right)^3 = \left(\frac{n}{2}\pm \frac{\operatorname{sign}(p + 3n^2)}{2}\sqrt{\frac{4p+3n^2}3}\right)^3 \\ = \left(\frac{n}{2}\pm \operatorname{sign}(p + 3n^2)\sqrt{\frac{4p+3n^2}{12}}\right)^3. \tag{6} $$
What can be said about $\epsilon$?
The only case in which $\epsilon = 0$ is the trivial case $(p,q) = (0,0)$. Then $n = 0$ is the only root of $(1)$ and thus $p + 3n^2 = 0$.
For $(p,q) \ne (0,0)$ we get the following:
If $\Delta = 4p + 3n^2 > 0$ (which means that $(1)$ has a single real root and two complex conjugate roots, as can be seen by considering the quadratic equation $x^2+nx+b = 0$), we conclude that also $p + 3n^2 > 0$ (if $p > 0$ it is trivial and if $p \le 0$ we have $p \ge 4p$). Therefore $\epsilon = \operatorname{sign}(p + 3n^2) = +1$.
If $\Delta = 0$ (which means that $(1)$ has a two real roots, one of them having multiplicity $2$), then any $\epsilon$ will do in $(5)$, thus $\epsilon$ is basically irrelevant. Nevertheless we prove that $\epsilon = +1$. Let us first observe that $n = 0$ is impossible. In fact, for $n = 0$ we get $q = 0^3 + p \cdot 0 + q = 0$ and $4p = 4p + 3n^2 = \Delta = 0$, thus $p = 0$ and hence $(p,q) = (0,0)$ which contradicts our above assumption. Since $n \ne 0$ we conclude that $p + 3n^2 > 0$ (if $p \ge 0$ it is trivial and if $p < 0$ we have $p + 3n^2 > 4p +3n^2 = \Delta = 0$). Therefore $\epsilon = \operatorname{sign}(p + 3n^2) = +1$.
If $\Delta < 0$ (which means that $(1)$ has three real roots), then it turns out that both $\epsilon = +1$ and $\epsilon = -1$ can occur as $\operatorname{sign}(p + 3n^2)$.
Finally, we can readily verify that
$$3\left(\frac{n}{2} + \frac{\epsilon}{2}\sqrt{\frac{\Delta}3}\right)\left(\frac{n}{2}- \frac{\epsilon}{2}\sqrt{\frac{\Delta}3}\right) + p = 0$$
which means that the solutions of $(1)$ are given via formula $(10)$ in Is there really analytic solution to cubic equation?
In particular we get
$$\frac{n}{2} + \frac{\epsilon}{2}\sqrt{\frac{\Delta}3} + \frac{n}{2}- \frac{\epsilon}{2}\sqrt{\frac{\Delta}3} = n$$
as a solution. This is hardly surprising because the above considerations were based on the fact that $n$ is a given root of $(1)$.
The real benefit of $(6)$ is that it helps to answer the question. In fact, if we have an expression of the form $\sqrt[3]{r \pm \sqrt{s}}$ with $r,s \in \mathbb Q$, we know that it occurs as one of the summands in the Cardano solution formula $(2)$ of the cubic equation $(3)$.
Let us now prove the theorem in the question. We consider $r, s \in \mathbb Q$ such that $\sqrt s \notin \mathbb Q$.
(a) Assume that equation $(3)$, i.e. $x^3 + 3\sqrt[3]{s - r^2} - 2r = 0$, has a rational root $n$.
Recall that this requires $\sqrt[3]{s - r^2} \in \mathbb Q$. Equation $(3)$ is a special case of $(1)$ with $p = 3\sqrt[3]{s - r^2}$ and $q = -2r$ which gives $-\frac{q}{2} \pm \sqrt{D} = r \pm \sqrt s$. Using $(6)$ we get
$$r \pm \sqrt s = \left(\frac{n}{2} \pm \epsilon \sqrt{\frac{12\sqrt[3]{s - r^2}+ 3 n^2}{12}}\right)^3 = \left(\frac{n}{2} \pm \epsilon \sqrt{\sqrt[3]{s - r^2}+ \frac{n^2}{4}}\right)^3 . \tag{7}$$
We have $\epsilon = \operatorname{sign}(p + 3n^2) = \operatorname{sign}(3\sqrt[3]{s - r^2} + 3n^2) = \operatorname{sign}(\sqrt[3]{s - r^2} + n^2)$. We shall see later (see point 6. below) that $\epsilon = +1$ when $s > 0$. If $s < 0$, then both $\epsilon = +1$ and $\epsilon = -1$ can occur. Summarizing, a rational type cubic root of $r \pm \sqrt s$ is given by
$$\rho_\pm(r,s,n) = \frac{n}{2} \pm \operatorname{sign}(\sqrt[3]{s - r^2} + n^2) \sqrt{\sqrt[3]{s - r^2}+ \frac{n^2}{4}}. \tag{8}$$
This formula shows that the number of rational roots of $(3)$ agrees with the number of rational type cubic roots of $r \pm \sqrt s$. Call this number $\nu(r,s)$. Clearly we may have $\nu(r,s) = 0$, but if $\nu(r,s) > 0$ we shall see later (see points 10 - 12. below) that $\nu(r,s) = 1$ unless $s = -3b^2$ for some $b \in \mathbb Q$ in which case $\nu(r,s) = 3$.
(b) Assume that $r + \sqrt s =(u + \epsilon \sqrt v)^3$ with $u,v \in \mathbb Q, \epsilon = \pm 1$.
We collect various properties of $u, v , \epsilon$.
$\sqrt v \notin \mathbb Q$.
Otherwise $r + \sqrt s \in \mathbb Q$, i.e. $\sqrt s \in \mathbb Q$.
$\sqrt v = c \sqrt s$ for some $c \in \mathbb Q$.
We have $r + \sqrt s = (u + \epsilon \sqrt v)^3 = u^3 + 3u^2 \epsilon \sqrt v + 3u v + \epsilon v \sqrt v = a + b \sqrt v$ with $a = u^3 + 3uv \in \mathbb Q$ and $b = \epsilon(3 u^2 + v) \in \mathbb Q$. It is impossible that $b = 0$ since in that case $\sqrt s = a - r \in \mathbb Q$. We get $\sqrt s - b \sqrt v = a - r$ and thus $s - 2b \sqrt s \sqrt v + b^2 v = (a-r)^2$, hence $\sqrt s \sqrt v \in \mathbb Q$. We conclude $\frac{\sqrt v}{\sqrt s} = \frac{\sqrt v \sqrt s}{s} = c \in \mathbb Q.$
If $s > 0$, then $v > 0$ and if $s < 0$, then $v < 0$.
It follows from 2. that both $\sqrt v, \sqrt s$ must be either real or purely imaginary.
$c > 0$.
By 2. we have $c = \frac{\sqrt v}{\sqrt s}$. Now we use 3. and our square root convention.
$r = u^3 + 3u v$, $1 = \epsilon (3u^2c + c^3s)$.
We have $$r + \sqrt s = (u + \epsilon c \sqrt s)^3 = u^3 + 3u^2 \epsilon c \sqrt s + 3u c^2s + \epsilon c^3 s \sqrt s \\= u^3 + 3u c^2s +\epsilon (3u^2c + c^3s)\sqrt s. $$ But $1$ and $\sqrt s$ are linearly independent over $\mathbb Q$ which implies 5.
If $s > 0$, then $\epsilon = +1$. If $s < 0$, then both $\epsilon = +1$ and $\epsilon = -1$ can occur.
We have $1 = \epsilon (3u^2c + c^3s)$. Since $3u^2c + c^3s > 0$ for $s > 0$, we must have $\epsilon = +1$. If $s < 0$ we see that both $\epsilon = +1$ and $\epsilon = -1$ are possible.
$u - \epsilon \sqrt v$ is a cubic root of $r - \sqrt s$.
This follows from 5.
$2u$ is a root of $(3)$. This finishes the proof of the theorem.
Case 1. $s < 0$.
By 3. $r + \sqrt s$ has the non-real cubic root $u + \epsilon \sqrt v$. Hence $2\operatorname{Re}(u + \epsilon \sqrt v) = 2u$ is a root of $(3)$. See the answer to Is there really analytic solution to cubic equation?
Case 2. $s > 0$.
We know that $u \pm \epsilon \sqrt v$ are the real roots of $r \pm \sqrt s$ and we conclude that $u + \epsilon \sqrt v + u - \epsilon \sqrt v = 2u$ is a root of $(3)$. See the answer to Is there really analytic solution to cubic equation?
We finally address the question whether rational type cubic roots are uniques. Let $\mathbb Q[\sqrt{-3}]$ denote the algebraic field extension of $\mathbb Q$ by $\sqrt{-3}$ which has $1,\sqrt{-3}$ as basis over $\mathbb Q$.
If $u + \epsilon \sqrt v = u' + \epsilon' \sqrt{v'}$ with $u,u', v, v' \in \mathbb Q$ and $\sqrt v, \sqrt{v'} \notin \mathbb Q$, then $u = u'$, $v = v'$ and $\epsilon = \epsilon'$. That is, each rational type cubic root has a unique representation.
We have $\epsilon \sqrt v - \epsilon' \sqrt{v'} = u' - u$, thus $v - 2\epsilon \epsilon' \sqrt v \sqrt{v'} + v' = (u' - u)^2$, thus $\sqrt v \sqrt{v'} \in \mathbb Q$. Hence $\sqrt{v'} = c \sqrt v$ for some $c \in \mathbb Q$. Thus both $\sqrt v, \sqrt{v'}$ must be either real or purely imaginary. Due to our square root convention we must have $c > 0$. But now $u + \epsilon \sqrt v = u' + \epsilon' c \sqrt{v}$ which implies $u = u'$ and $\epsilon = \epsilon' c$. Since $c >0$, we get $\epsilon = \epsilon'$ and $c = 1$. Thus $\sqrt{v'} = \sqrt v$ which shows that $v' = v$.
If $\sqrt s \notin Q[\sqrt{-3}]$ and $u + \epsilon \sqrt v, u' + \epsilon' \sqrt{v'}$ are rational type cubic roots of $r + \sqrt s$, then $u = u'$, $v = v'$ and $\epsilon = \epsilon'$.
We know that $\sqrt v = c \sqrt s, \sqrt{v'} = c'\sqrt s$ for suitable positive $c, c' \in \mathbb Q$. We shall show that $u + \epsilon c \sqrt s = u' + \epsilon' c' \sqrt s$ which implies $u = u'$ and $\epsilon c = \epsilon' c'$. Since $c, c' > 0$ we conclude $\epsilon = \epsilon'$ and $c = c'$. Thus $\sqrt v = \sqrt{v'}$ and therefore $v = v'$.
Assume that $u + \epsilon c \sqrt s \ne u' + \epsilon' c' \sqrt s$. Then the quotient $\frac{u' + \epsilon' c'\sqrt s}{u + \epsilon c \sqrt s}$ must be a non-trivial third cubic unit root. Thus $u' + \epsilon' c' \sqrt s = (u + \epsilon c \sqrt s)(\frac{1\pm \sqrt{-3}}{2})$. This gives $2u' + 2\epsilon' c'\sqrt s = u \pm u \sqrt{-3} + \epsilon c \sqrt s \pm \epsilon c \sqrt s \sqrt{-3}$, i.e. $(2\epsilon' c' - \epsilon c \mp \epsilon c \sqrt{-3})\sqrt s = u \pm u \sqrt{-3} - 2u'$. Since $\epsilon c \ne 0$, we have $2\epsilon' c' - \epsilon c \mp \epsilon c \sqrt{-3} \ne 0$ and therefore $\sqrt s \in \mathbb Q[\sqrt{-3}]$ which is a contradiction.
$\sqrt s \in Q[\sqrt{-3}]$ if and only $s = -3b^2$ for some $b \in \mathbb Q$.
If $\sqrt s \in Q[\sqrt{-3}]$, then $\sqrt s = a + b\sqrt{-3}$ with $a, b \in \mathbb Q$. This requires $b \ne 0$. We get $s = a^2 + 2ab \sqrt{-3} - 3b^2$ which is only possible when $2ab = 0$. Therefore $a = 0$. The converse is obvious.
If $s = -3b^2$ for some $b \in \mathbb Q$ and $r + \sqrt s$ has a rational type cubic root $u + \epsilon \sqrt v$, then there are exactly two more rational type cubic roots. These are $w_\pm = (u + \epsilon \sqrt v)(\frac{1\pm \sqrt{-3}}{2})$ which have again the form $u' + \epsilon' \sqrt{v'}$.
We can write $\sqrt v = c \sqrt s = d \sqrt{-3}$ with some $d \in \mathbb Q$. Then $w_\pm = \alpha_\pm + \beta_\pm \sqrt{-3} = \alpha_\pm + \epsilon_\pm \sqrt{-3\beta_\pm^2}$ with suitable $\alpha_\pm, \beta_\pm \in \mathbb Q$.
Let us study a few examples.
$1 + \sqrt 3$ has no rational type cubic root because $\sqrt[3]{3 - 1^2} = \sqrt[3] 2 \notin \mathbb Q$.
$1 + \sqrt 2$ has no rational type cubic root. We have $\sqrt[3]{2 - 1^2} = 1 \in \mathbb Q$, but $x^3 + 3x - 2 = 0$ does not have a rational root.
$1 +\sqrt{\frac{28}{27}}$ has a rational type cubic root. In fact, $\sqrt[3]{\frac{28}{27} - 1^2} = \sqrt[3]\frac{1}{27} = \frac{1}{3}$ and $(3)$ gets
$$x^3 + x -2 = 0 .$$
It has $n = 1$ as rational root and we get the rational type cubic root
$$\frac{1}{2} + \sqrt{\frac{1}{3} + \frac{1}{4}} = \frac{1}{2} + \sqrt{\frac{7}{12}} .$$
$45 + 29\sqrt 2 = 45 + \sqrt{1682}$ has a rational type cubic root. In fact, $\sqrt[3]{1682 - 45^2} = \sqrt[3]{-343} = -7$ and $(3)$ gets
$$ x^3 - 21x -90 = 0 .$$
It has $n = 6$ as rational root and we get the rational type cubic root
$$3 + \sqrt{-7 + \frac{6^2}{4}} = 3 + \sqrt{2} .$$
$2 + \sqrt{-121}$ has a rational type cubic root. In fact, $\sqrt[3]{-121 - 2^2} = \sqrt[3]{-125} = -5$ and $(3)$ gets
$$x^3 -15x - 4 = 0 .$$
It has $n = 4$ as rational root and we get $\epsilon = \operatorname{sign}(-5 + 4^2)
= +1$ and thus
$$2 + \sqrt{-5 + \frac{4^2}{4}} = 2 + \sqrt{-1}$$
is a rational type cubic root. Note that Rafael Bombelli (1526 - 1572) used this to solve the equation $x^3 = 15x + 4$ via the Cardano formula. I guess he found this cubic root by trying.
$-3 + \frac{10}{9}\sqrt{-3} = -3 + \sqrt{-\frac{100}{27}}$ has a rational type cubic root. In fact, $\sqrt[3]{-\frac{100}{27} - (-3)^2} = \sqrt[3]{-\frac{343}{27}} = -\frac{7}{3}$ and $(3)$ gets
$$x^3 -7x +6 = 0 .$$
It has $n = 1$ as rational root and we get $\epsilon = \operatorname{sign}(-7 + 1^2) = -1$ and thus
$$\frac{1}{2} - \sqrt{-\frac{7}{3} + \frac{1^2}{4}} = \frac{1}{2} - \sqrt{\frac{-25}{12}} = \frac{1}{2} - \frac{5}{2} \sqrt{\frac{-1}{3}}$$
is a rational type cubic root.
Update:
The above considerations also show the following for a cubic equation $x^3 + px + q = 0$ with $p,q \in \mathbb Q$:
Let $x^3 + px + q = 0$ have a rational root. Then there exist $u, v \in \mathbb Q$ and $\epsilon = \pm 1$ such that $u \pm \epsilon \sqrt v$ is a a cubic root of $-\frac{q}{2} \pm \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}$. See $(6)$.
Let $t = -\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}$ have a cubic root of the form $u \pm \sqrt v$ with $u, v \in \mathbb Q$. If $t \ne 0$, then $x^3 + px + q = 0$ has a rational root. Note that the case $t = 0$ occurs precisely when $p = 0$ and $q \ge 0$. Thus we consider the trvial case of an equation $x^3 + q = 0$. This has a rational root if and only if the real cubic root of $q$ is rational.
Case 1. $\sqrt{\frac{q^2}{4} + \frac{p^3}{27}} \notin \mathbb Q$. Then we conclude that $t \ne 0$. Moreover, our above considerations show that $2u$ is a rational root of $x^3 + px + q = 0$.
Case 2. $\sqrt{\frac{q^2}{4} + \frac{p^3}{27}} \in \mathbb Q$. Then $t \in \mathbb Q$ and the argument at the end of the question shows that $t$ has a rational cubic root $w_+$. Since $t \ne 0$, we get $w_+ \ne 0$ and thus also $w_- = -\frac{p}{3w_+} \in \mathbb Q$. But then the answer to Is there really analytic solution to cubic equation? shows that $w_+ + w_-$ is a rational root of $x^3 + px + q = 0$.
Best Answer
In your question and comment there are a few different things that might warrant addressing. I'll try to respond to them separately, starting with some minor points of clarification, and then answering the closest thing to what you want.
Unfortunately, your work seems to be looking for when the "first $t$" is positive in the solution of the depressed cubic equation $t^3+pt+q=0$. But since $x=t-\frac{b}{3a}$, answering that question is not the same as answering when $x$ is positive, because of the shift by $\dfrac{b}{3a}$. (It turns out that when the equation has three real roots, the $\frac{b}{3a}$ shift makes the biggest $t$ root positive no matter what.)
Unfortunately, there's a significant issue with this idea. When there's exactly one real root, it's $x_1$, and then you're asking "is the real root positive?", which is a very sensible question. But when there are more real roots, then you're taking cube roots of non-real numbers (and you can't really avoid it: you're likely in the casus irreducibilis), which sort of depends on how you how you define them.
Now, if you arbitrarily pick a particular definition of complex cube root, you can answer this, or if you arbitrarily pick an ordering of the trigonometric formulas for the solutions then you have a concrete question. But since you didn't mention either of those approaches in your question I'm going to assume that you didn't really have them in mind. If you're interested in something like that, I suppose you could ask a separate question about it.
The real question:
Two key facts that can help us answer your question.
What do these two facts tell us?
What are all the cases, though?
For simplicity, divide through your polynomial $ax^3+bx^2+cx+d$ by the leading coefficient $a$ to give the the form $x^3+bx^2+cx+d$. The roots will be the same, but now we have cut the cases in half since we don't have to worry about $a$ being negative. (The discriminant is now a little simpler, too: $\Delta=-4 b^3 d+b^2 c^2+18 b c d-4 c^3-27 d^2$.)
Now we can make a table of possibilities for the signs of the coefficients. I'll use inequality symbols as shorthand for comparison with $0$, so that $\ge$ will mean $\ge0$ in this table. The numbers of sign changes come from that fact that the leading coefficient has been made $1$. The last column is the number of positive roots.
\begin{array}{cccc} b & c & d & \text{#}\\ \ge & \ge & \ge & 0\\ \ge & \ge & < & 1\\ \ge & < & \le & 1\\ \ge & < & > & 0\text{ or }2\\ < & \le & \le & 1\\ < & \le & > & 0\text{ or }2\\ < & > & \ge & 0\text{ or }2\\ < & > & < & 1\text{ or }3 \end{array} And from our notes above, every single one of those "or" cases is distinguished by the discriminant: If $\Delta>0$ then it's the lower number, and if $\Delta\le0$ then it must be the higher number.
Closing remarks
The discriminant is kind of an ugly expression, and this table has a bunch of cases, but unfortunately there's not much you can do about that. You can regroup the cases a bit if you like, but I don't think it can really be reduced in general. One general rewriting that still has 8 cases is: \begin{array}{cccc} b & c & d & \text{#}\\ \ge & \ge & \ge & 0\\ \ge & ? & < & 1\\ ? & < & = & 1\\ < & \le & \le & 1\\ \ge & < & > & 0\text{ or }2\\ < & ? & > & 0\text{ or }2\\ < & > & = & 0\text{ or }2\\ < & > & < & 1\text{ or }3 \end{array}
That said, if you happen to know that $\Delta>0$ so that there's only one real root (and hence it's $x_1$), then things do simplify somewhat nicely: \begin{array}{cccc} b & c & d & x_1>0\text{?}\\ ? & ? & > & \text{no}\\ ? & > & = & \text{no}\\ \ge & = & = & \text{no}\\ ? & ? & < & \text{yes}\\ ? & < & = & \text{yes}\\ < & = & = & \text{yes} \end{array}