Thanks to those who answered. Doing more research, while browsing the book "Generic Polynomials" (thanks, Rouse!), in page 30 I saw the generic cubic for $C_3 = A_3$,
$$x^3 + m x^2 - (m + 3)x + 1 = 0$$
which has negative square discriminant $D = -(m^2+3m+9)^2,$ hence all roots $x_i$ are real. After some experimentation, if we define,
$$a = x_1^{1/3},\quad b = x_2^{1/3},\quad c = x_3^{1/3}$$
then we have,
$$\color{blue}{a^2b+b^2c+c^2a = 0}$$ $$a^2c+b^2a+c^2b \neq 0$$
so the correct order of roots $x_i$ of the generic cubic must be chosen. IF TRUE, this explains the properties below for levels $9, 13, 27$.
I. Level 9
The level $9$ theta quotients satisfy the cubic,
$$x^3+(u+3)x^2-(u+6)x+1=0$$
which, after minor change of variables, is just the generic cubic in disguise. With $u = \left(\frac{\eta(\tau)}{\eta(9\tau)}\right)^3$, the roots are,
$$x_1 = \frac{\eta(3\tau)}{\eta(\tau)} \left(\frac{\eta(3\tau)}{\eta(9\tau)}\right)^3 \left(\frac{q^{5/9}\,f(-q,-q^8)}{\quad f(-q^3)}\right)^3$$
$$x_2 = \frac{\eta(3\tau)}{\eta(\tau)} \left(\frac{\eta(3\tau)}{\eta(9\tau)}\right)^3 \left(\frac{q^{2/9}\,f(-q^2,-q^7)}{\quad f(-q^3)}\right)^3$$
$$x_3 = \frac{\eta(3\tau)}{\eta(\tau)} \left(\frac{\eta(3\tau)}{\eta(9\tau)}\right)^3 \left(\frac{-\,f(-q^4,-q^5)}{q^{1/9}\,f(-q^3)\;}\right)^3$$
with $f(a,b)$ the usual Ramanujan theta function. Taking their cube roots $\sqrt[3]{x_i}$, they then obey the equation in blue, a relation mentioned in the original post above.
II. Level 13
As discovered by Ramanujan, the level $13$ theta quotients satisfy the cubic,
$$x^3+(v+1)x^2-(v+4)x+1=0$$
which, after minor change of variables, is also the generic cubic in disguise. With $v = \left(\frac{\eta(\tau)}{\eta(13\tau)}\right)^2$, the roots are,
$$x_1 = \left(\frac{f(-q^2,-q^{11})}{f(-q,-q^{12})}\right) \left(\frac{f(-q^{10},-q^{3})}{f(-q^5,-q^{8})}\right)$$
$$x_2 = \frac{-1}{\,q}\left(\frac{f(-q^4,-q^{9})}{f(-q^2,-q^{11})}\right) \left(\frac{f(-q^{6},-q^{7})}{f(-q^3,-q^{10})}\right)$$
$$x_3 = \frac{1}{q^{-1}}\left(\frac{f(-q^8,-q^{5})}{f(-q^4,-q^{9})}\right) \left(\frac{f(-q^{12},-q)}{f(-q^6,-q^{7})}\right)$$
Taking their cube roots $\sqrt[3]{x_i}$, they obey the equation in blue.
III. Level 27
The Dyson Mod 27 Identities also apply, at least three of them. (Caveat: I changed the order in the Mathworld link.) Those satisfy the cubic,
$$x^3+(w+3)x^2-(w+6)x+1=0$$
which is just a variation for level 9 but with $w = \left(\frac{\eta(3\tau)}{\eta(27\tau)}\right)^3$. The roots are,
$$x_1 = \frac{\eta(9\tau)}{\eta(3\tau)} \left(\frac{\eta(\tau)}{\eta(27\tau)}\right)^3 \left(\frac{q^{2}\,f(-q^3,-q^{24})}{\quad f(-q)}\right)^3$$
$$x_2 = \frac{\eta(9\tau)}{\eta(3\tau)} \left(\frac{\eta(\tau)}{\eta(27\tau)}\right)^3 \left(\frac{q\,f(-q^6,-q^{21})}{\quad f(-q)}\right)^3$$
$$x_3 = \frac{\eta(9\tau)}{\eta(3\tau)} \left(\frac{\eta(\tau)}{\eta(27\tau)}\right)^3 \left(\frac{-\,f(-q^{12},-q^{15})}{f(-q)\;}\right)^3$$
Taking their cube roots $\sqrt[3]{x_i}$, they also obey the equation in blue. (I've ignored the fourth Dyson Mod 27 Identity, so a more satisfying relation is a quartic with coefficients determined by some eta quotient.)
IV. Conclusion
Just like the Klein quartic $a^3b+b^3c+c^3a$, it seems there is more to,
$$a^2b + b^2c+ c^2a = 0$$
than meets the eye. Does it have a name, so it is Internet searchable?