Thanks to those who answered. Doing more research, while browsing the book ["Generic Polynomials"][1] (*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.

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**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, they then obey the equation in blue, a relation mentioned in the original post above.

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**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, they also obey the equation in blue.

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**III. Level 27**

The [Dyson Mod 27 Identities][2] also apply, at least three of them. (*Caveat*: I changed the order in the Mathworld link.) Define,

$$a = q^2\;  \frac{f(-q^3,-q^{24})}{f(-q)}$$

$$b = q\;  \frac{f(-q^6,-q^{21})}{f(-q)}$$  

$$c = - \frac{f(-q^{12},-q^{15})}{f(-q)}$$

$$d = q^{1/3}\;  \frac{f(-q^9,-q^{18})}{f(-q)}$$

So again $a,b,c$ obey the equation in blue. (The $d$ version for level $9$ degenerates to $d=1$.) The cubes of $a,b,c$ must then satisfy a cubic similar to the generic cubic. 

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**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. But was it ever given a name?

  [1]: http://library.msri.org/books/Book45/files/book45.pdf
  [2]: https://mathworld.wolfram.com/DysonMod27Identities.html