While @dxiv has essentially answered the question, we can add some more details for future posts, plus a cameo by the Bring-Jerrard quintic.
I. Deriving formulas
First, if one has an expression of form,
$$Z = -X + \sqrt[m]{a+\sqrt[n]b}+\sqrt[m]{a-\sqrt[n]b}$$
move the bothersome $X$ out of the way by assuming a new variable $\color{blue}{Z= T-X}$ to get the simpler,
$$T = \sqrt[m]{a+\sqrt[n]b}+\sqrt[m]{a-\sqrt[n]b}$$
Once the minimal polynomial for $T$ has been found, it is easy to re-express it in terms of $Z$ using the blue equation.
Second, yes, given one root of any equation, all of its Galois conjugates can be found. The beauty of Galois theory is it unified all known and "future" formulas into a single framework. All equations can be transformed into "depressed" form and, if prime degree $p$ with a solvable Galois group, have predictable formulas. For $p = 2,3,5$, we have
$$z_k = u^{1/2}\,\zeta_2^k$$
$$z_k = u_1^{1/3}\,\zeta_3^k+u_2^{1/3}\,\zeta_3^{2k}$$
$$z_k = u_1^{1/5}\,\zeta_5^k+u_2^{1/5}\,\zeta_5^{2k}+ u_3^{1/5}\,\zeta_5^{3k}+u_4^{1/5}\,\zeta_5^{4k}\\$$
respectively, with roots of unity $\zeta_p = e^{2\pi i/p}$, the $u_i$ as roots of the resolvent equation of degree $p-1$, and $k = (0 \text{ to } p-1),$ implying the equation of degree $p$ has $p$ roots. (Your cubic root was just the case $k=0$.) For example, for $p=3$, then,
$$P(z)=(z-z_0)(z-z_1)(z-z_2) = 0$$
which simplifies to,
$$\color{brown}{P(z)=z^3-3(u_1u_2)^{1/3}z-(u_1+u_2) =0}$$
For cubics, its resolvent is a quadratic, and we have the OP's,
$$u_i = Y \pm \sqrt{Y^2 - \frac{X^6}{27}}$$
So using the $u_i$ on the brown equation, it yields,
\begin{align}P(z)
&= z^3-3\Big(\tfrac{X^6}{27}\Big)^{1/3}z-(2Y)=0\\
&= z^3-X^2z-2Y= 0
\end{align}
Let $z = Z+X$ and we recover,
$$Z^3+3XZ^2+2X^2Z−2Y = 0$$
$$Z(Z + 2X)(Z+X) − 2 Y = 0$$
II. Second family and Bring-Jerrard quintic
Per the OP's formula, the first family is,
- $f_0(Z) = Z (Z + 1 X)^0 − 1 Y$
- $f_1(Z) = Z (Z + 2 X)^1 − 1 Y$
- $f_2(Z) = Z (Z + 3 X)^2 − 2 Y$
- $f_3(Z) = Z (Z + 4 X)^3 − 8 Y$
- $f_4(Z) = Z (Z + 5 X)^4 − 64 Y$
and so on. The nice thing about this family is that, for any $n$, it has a closed-form solution in terms of generalized hypergeometric functions. Given the more general form,
$$Z (Z + (n+1) X)^n − W =0$$
Let $Z=z^n,$ $X=a,$ $W=(nb)^n,$ thus,
$$\big(z^{n+1} + (n+1)az\big)^n − \big(nb\big)^n =0$$
This is a difference of two $n$th powers and can be factored. We get the trinomial factor,
$$z^{n+1}+(n+1)az-nb = 0\;$$
For example, for $n=4$, yields the Bring-Jerrard quintic,
$$z^5+5az-4b = 0$$
and a root $z$ is,
$$z = \frac{4b}{5a}\,{_4F_3}\Big(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,-\frac{b^4}{a^5}\Big)$$
Then we have $Z = z^n$ by undoing all the substitutions. And so on for any $n$.
III. Second family
The second family seems to be,
- $g_0(Z) = \; ??$
- $g_1(Z) = Z(Z + 2X) − 1 Y$
- $g_2(Z) = Z(Z + 2X)(Z+X) − 2 Y$
- $g_3(Z) = Z(Z + 2X)(Z+\color{blue}aX)(Z+\color{blue}bX) − \color{blue}cY$
for some small integer $a,b,c$, though it hard to ascertain those without more details.
Best Answer
Consistent with the comment of dxiv,
let $~y = x-1,~$ and let $~u = y^2.$
Then
$$(y^2 - 4)(y^2 - 36) = 1680 \implies $$
$$u^2 - 40u - 1536 = 0 \implies \tag1 $$
$$u = \frac{1}{2} ~\left[40 \pm \sqrt{1600 + (4 \times 1536)}\right] \tag2 $$
$$ = \frac{1}{2} \left[40 \pm 88\right] = 20 \pm 44.$$
Since you are looking for real solutions, you must have that
$$y^2 = u = 64 \implies (x-1) = y = \pm 8.$$
Addendum
Actually, all of the Math, following (2) above is unnecessary. From (2) above, it is immediate that the value inside the radical will be bigger than $~(40)^2,~$ and so only one (positive) value of $~y^2 = u~$ will be forthcoming.
Then regardless of the eventual computed value of $~u = y^2,~$ you will have the two real roots of
$$x-1 = y = \pm \sqrt{u}.$$
So, denoting the two roots as $~x_1,~$ and $~x_2,~$ you have that $~\displaystyle (x_1 - 1) + (x_2 - 1) = \sqrt{u} + (-\sqrt{u}) = 0.$
So, without bothering to solve for $~u,~$ you know that $~x_1 + x_2 = 2.$
In fact, with experience, you can see that because the third term in (1) above, $~-1536,~$ is negative, you could (instead) stop the Math after (1) above, rather than (2) above.
That is, (1) above, with any negative 3rd term, is sufficient to imply that only one positive real root of $~y^2 = u~$ will be forthcoming.