The OP asks for any feedback. In fact, the simple-looking cubic,
$$(x+\beta)^3 = \alpha x^2\tag1$$
has a LOT of interesting properties, being connected to Ramanujan's theta function, the Klein quartic curve, and septic equations. First, some minor transformations. Let $x = 1/y,\; \alpha=n^2+13n+49,$ and $\beta=1$ to get,
$$y^3+3y^2-(n^2+13n+46)y+1 = 0\tag{2a}$$
This retains the two properties mentioned by the OP, plus some additional ones. Given the Ramanujan theta function $f(a,b)$, Dedekind eta function $\eta(\tau)$, and $q=e^{2\pi i\tau}$, define,
\begin{align}
a &= -q^{61/168}\,\frac{f(-q,-q^6)}{f(-q^2)}\\
b &= \; q^{13/168}\;\frac{f(-q^2,-q^5)}{f(-q^2)}\\
c &= q^{-11/168}\,\frac{f(-q^3,-q^4)}{f(-q^2)}\\
n &= \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4\\
\psi &= \left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(7\tau)}\right)^4
\end{align}
such that $(a,b,c)$ obey the Klein quartic curve,
$$a^3b+b^3c+c^3a = 0$$
Then the roots $y_k$ of the cubic have the form,
\begin{align}
y_1 &= -(a^3b)^3\,\psi\\
y_2 &= -(b^3c)^3\,\psi\\
y_3 &= -(c^3a)^3\,\psi
\end{align}
which explains why the roots of,
$$y^3+3y^2-(n^2+13n+46)y+1 = 0\tag{2b}$$
even for any $n$ have,
$$y_1^{1/3} + y_2^{1/3} + y_3^{1/3} = 0$$
since they are just obeying the Klein quartic curve. However, if we take 7th roots instead,
$$z=-(y_1^3\,y_2)^{1/7}-(y_2^3\,y_3)^{1/7}-(y_3^3\,y_1)^{1/7}\tag3$$
then $z$ is a solution in radicals to the solvable septic,
$$z^7 + 14 z^4 - 21 z^3 - 14 (46 + 13 n + n^2) z^2 - 7 (249 + 114 n + 18 n^2 + n^3) z - (1348 + 854 n + 207 n^2 + 23 n^3 + n^4) = 0$$
For example, let $n=-6$, then the septic,
$$z^7 + 14 z^4 - 21 z^3 - 56 z^2 + 21 z - 4 = 0$$
is solvable by the cubic,
$$y^3 + 3 y^2 - 4 y + 1 = 0$$
using the relation $(3)$ and the roots $y_k$ correctly ordered. There's more, but this is enough. And to think all these inter-connections came from the simple structure $(x+\beta)^3 = \alpha x^2.$
Addendum: Since the OP asked about the special functions $(a,b,c)$ of order 7 (more on these functions in this MO post), they are generalizations of the Roger-Ramanujan identities of order 5. Define,
\begin{align}
u &=\, q^{11/60}\,\frac{f(-q,-q^4)}{f(-q)}\; =\; q^{11/60}\prod_{n=1}^\infty \frac1{\big(1-q^{5n-2}\big)\big(1-q^{5n-3}\big)}\\
v &= q^{-1/60}\,\frac{f(-q^2,-q^3)}{f(-q)} = q^{-1/60}\prod_{n=1}^\infty \frac1{\big(1-q^{5n-1}\big)\big(1-q^{5n-4}\big)}
\end{align}
which now is related to the quadratic,
$$x^2-(m+11)x-1=0$$
with roots $(x_1, x_2) = (-u^{10}m^5,\;v^{10}m^5)$ and $m = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6.$
Radicals
The eta quotients $(m,n,\psi)$ are radicals when $\tau = \sqrt{-d}$, which implies the special functions $(u,v)$ and $(a,b,c)$ are radicals are well. For example, let $d=1$, then,
\begin{align}
m &= \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6 = 5^3\left(\frac{1+\sqrt5}{2}\right)^3\\
n &= \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4 =\frac{28\times7^{3/4}\sqrt2}{\left(\sqrt{5-\sqrt7}-\sqrt{-7+3\sqrt7}\right)^2} \\
\psi &= \left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(7\tau)}\right)^4 =\frac{7^{3}\sqrt{14}}{\left(\sqrt{5-\sqrt7}-\sqrt{-7+3\sqrt7}\right)^4} \\
\end{align}
Substituting these into the quadratic and cubic, then solving for their roots, one can then solve for $(u,v)$ and $(a,b,c)$ in terms of radicals.
Multiplying by $18$ and rearranging, the equality to be proved is:
$$
18(a^2-b) x y + 9(a b-c)(x+y) \pm 9\delta(x - y) + 18(b^2 - a c)=0 \tag{1}
$$
Let the third root be $z$, then the elementary symmetric polynomials in $x,y$ can be written in terms of $z$ as:
$$
\begin{align}
x+y = -3a-z \tag{2}
\\ xy = -\frac{c}{z} \tag{3}
\end{align}
$$
Choosing the sign for $\pm \sqrt{\Delta}$ such that $\delta = \frac{1}{9}(x-y)(y-z)(z-x)\,$:
$$
\begin{align}
-9\delta(x-y) &= (x-y)^2(z-x)(z-y)
\\ &= \big((x+y)^2 - 4xy\big)(z^2 - (x+y)z + xy)
\\ &= \big(z^2 + 6a z + 9a^2 + \frac{4c}{z}\big)\big(2z^2 + 3 a z - \frac{c}{z}\big)
\\ &= \frac{1}{z^2}(z^3+6az^2+9a^2z+4c)(2z^3 + 3 a z^2 - c) \tag{4}
\end{align}
$$
$\big($Another way to read $(4)$ is as an expression for $\,|x-y|=\dfrac{\pm\,9\delta z}{2z^3 + 3 a z^2 - c}\,$ in terms of $\,z, \delta\,$.$\big)$
Substituting $(2),(3),(4)$ in $(1)\,$:
$$
- 18(a^2-b) \frac{c}{z} - 9(a b-c)(3a + z) - \frac{1}{z^2}(z^3+6az^2+9a^2z+4c)(2z^3+3az^2-c) + 18(b^2 - a c) = 0
$$
Clearing the denominators gives a sextic in $z$, which has the original cubic as a factor (verified by WA):
$$
\begin{align}
&- 18(a^2-b) c z - 9(a b-c)(3a + z)z^2 - (z^3+6az^2+9a^2z+4c)(2z^3+3az^2-c) + 18(b^2 - a c)z^2
\\ &\;\;=\;\; -(z^3 + 3 a z^2 + 3 b z + c) \left(2 z^3 + 9 a z^2 + 3(3 a^2 - 2 b) z - 4 c\right)
\\ &\;\;=\;\; 0
\end{align}
$$
Best Answer
Let $f=3t$ then the cubic is:
$$ p(x) = (x+y)^3 - x^2 z + t x z (3 x + 3 y + t^2 z) = \prod (x-x_k) $$
Let $\omega$ be a primitive cube root of unity, and $q(x) = x^3 - u x^2 + v x - w$ the cubic with roots $\sqrt[3]{x_k}$, then:
$$ \begin{align} q(x)\,q(\omega x)\,q(\omega^2x) &= \prod (x-\sqrt[3]{x_k})(\omega x-\sqrt[3]{x_k})(\omega^2x-\sqrt[3]{x_k}) \\ &= \prod (x^3 - x_k) \\ &= p(x^3) \end{align} $$
It follows that:
$$ \begin{align} p(x^3) &= (x^3 - u x^2 + v x - w)(x^3 - \omega^2\,u x^2 + \omega\, v x - w)(x^3 - \omega\,u x^2 + \omega^2\,v x - w) \\ &= \color{blue}{x^9} - x^6 (u^3 - 3 u v + \color{blue}{3 w}) + x^3 (-3 u v w + v^3 + \color{blue}{3 w^2}) - \color{blue}{w^3} \\ &= \color{blue}{(x^3-w)^3} - x^6 u^3 + x^3 \left(3 u v x^3 - 3u v w + v^3\right) \end{align} $$
By inspection, comparing to $\,p(x^3) = (x^3+y)^3 - x^6 z + x^3 (3 t z x^3 + 3 t z y + t^3 z^2)\,$:
$$ \begin{cases} \begin{align} u &= \sqrt[3]{z} \\ v &= t \sqrt[3]{z^2} \\ w &= -y \end{align} \end{cases} $$
Therefore $\,q(x)=x^3 - \sqrt[3]{z}\,x^2 + t\sqrt[3]{z^2}\,x + y\,$, so $\,\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3} = u = \sqrt[3]{z}\,$.
(Note: just for reference, below is the first and very different revision of my answer.)
To get this out the way while looking forward to a more satisfying answer, the following is a brute force breakdown. With $f=3t$ the cubic is:
$$(x+y)^3 - x^2 z + t x z ( 3 x + 3 y + t^2 z) \;=\; x^3 + x^2 (3 y + 3 t z - z) + x (3 y^2 + 3 t y z + t^3 z^2) + y^3$$
@River Li's answer here provides a formula for $\,\sum \sqrt[3]{x_k}\,$ for generic cubics with distinct real roots (the assumption of rational coefficients is not needed or used AFAICT).
With those notations, the cubic here has:
$$ \begin{align} b &= 3 y + 3 t z - z \\ c &= 3 y^2 + 3 t y z + t^3 z^2 \\ d &= y \\ Q &= 4 (t z + 3 y) (3 t^3 z^2 - t^2 z^2 + 12 t y z + 18 y^2 - 3 y z) \\ \Delta &= -z^2 ((t - 1) t^2 z- y)^2 (t^2 (4 t - 1) z^2 + 2 (9 t - 2) y z + 27 y^2) \\ Q^2 - 16 \Delta &= \big(4 (t^3 z^2 + (6 t - 1) y z + 9 y^2)\big)^3 \end{align} $$
With the notations from my answer here about denesting cube roots like $\,\sqrt[3]{Q\pm4\sqrt{\Delta}}\,$:
$$ \begin{align} r &= 4 (t^3 z^2 + (6 t - 1) y z + 9 y^2) \\ s &= - t z - 3 y \\ \sqrt[3]{Q+4\sqrt{\Delta}} + \sqrt[3]{Q-4\sqrt{\Delta}} = 2s &= -2(t z + 3 y) \end{align} $$
It follows that:
$$ {-\frac{3}{2}\Big(\sqrt[3]{Q+4\sqrt{\Delta}}+\sqrt[3]{Q-4\sqrt{\Delta}}\Big) -b-6d} = 3(t z + 3 y) - (3 y + 3 t z - z) -6y = z $$
Which by $\,(*)\,$ concludes that $\,\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3} = \sqrt[3]{z}\,$.