I have been looking at binary quadratic forms for a question on MSE, number 3985889, about forms representing a prime (not dividing the discriminant) and primitively representing its cube. I calculated that the order of such a form must be one or two or four. The group under Gauss composition, mostly using Dirichlet's description.

When such a form has nice coefficients as $f=ax^2 + mc xy + ac y^2,$ triple $\langle a,mc,ac \rangle,$ some pleasant things happen. The duplicate form has coefficients $\langle a^2,mc,c \rangle$ and the fourth power is evidently the identity, as $c | mc.$ This is what Dickson calls an ambiguous representative of the class. We define $ X =-ax^3 + 3acx y^2 + mc^2 y^3 \, , \; \; \; $ $ Y = mx^3 + 3ax^2 y -ac y^3 \, , \; \; \; $ after which $ F = a X^2 + mcXY + ac Y^2 $ is identically equal to $f^3$ as polynomials in $x,y.$

I have a single strange example so far, $\langle 14, 8, 29 \rangle.$ It is of order four, and with $ f = 14 x^2 + 8 xy+29y^2, $ $ u= 6x^3 + 60x^2y - 3xy^2 -42y^3, $ $ v = 8x^3- 18x^2y - 60xy^2 +y^3,$ $ h = 14 u^2 + 8 uv+29v^2 $ cause $h=f^3$ identically as polynomials.

However, I was unable to put the form $\langle 14, 8, 29 \rangle$ into the desired shape $\langle a,mc,ac \rangle.$ For a favorable shape, the duplicate $\langle 10, 0, 39 \rangle$ would need to have a representative either $\langle 10, 20v, 39 + 10 v^2 \rangle$ or $\langle 39, 78v, 10 + 39 v^2 \rangle$ where the final coefficient is to be a square, in particular the square of something represented by $\langle 14, 8, 29 \rangle.$ But that does not happen; the proof involves a half dozen Pell type equations.

So there are the questions: why does $\langle 14, 8, 29 \rangle$ have no equivalent expression as $\langle a,mc,ac \rangle,$ and where might we find other examples?