I should have stuck with your preferred notation, as in your $B^2 + B C - 57 C^2 = A^3$ in a comment. So the form of interest will be $x^2 + x y - 57 y^2.$The other classes with this discriminant of indefinite integral binary quadratic forms would then be given by
$ 3 x^2 \pm xy - 19 y^2.$

Therefore, take
$$ \phi(x,y) = x^2 + x y - 57 y^2.$$
The identity you need to deal with your $A= \pm 3$ is
$$ \phi( 15 x^3 - 99 x^2 y + 252 x y^2 - 181 y^3 , \; 2 x^3 - 15 x^2 y + 33 x y^2 - 28 y^3 ) \; = \; ( 3 x^2 + xy - 19 y^2 )^3 $$

This leads most directly to $\phi(15,2) = 27.$ Using $ 3 x^2 + x y - 19 y^2 = -3$ when $x=7, y=3,$ this leads directly to $ \phi(1581, -196) = -27.$

However, we have an automorph of $\phi,$
$$ W \; = \;
\left( \begin{array}{rr}
106 & 855 \\\
15 & 121
\end{array}
\right) ,
$$
and
$ W \cdot (1581,-196)^T = (6, -1)^T,$ so $\phi(6,-1) = -27.$

Finally, any principal form of odd discriminant, call it $x^2 + x y + k y^2,$ (you have $k=-57$) has the improper automorph
$$ Z \; = \;
\left( \begin{array}{rr}
1 & 1 \\\
0 & -1
\end{array}
\right) ,
$$
while $ Z \cdot (6,-1)^T = (5, 1)^T,$ so $\phi(5,1) = -27.$

EDIT: a single formula cannot be visually obvious for all desired outcomes. There are an infinite number of integral solutions to $3 x^2 + x y - 19y^2 = -3.$ It is an excellent bet that one of these leads, through the identity I give, to at least one of the desired $\phi(5,1) = -27$ or $\phi(6,-1) = -27,$ but not necessarily both, largely because
$3 x^2 + x y - 19y^2$ and $3 x^2 - x y - 19y^2$ are not properly equivalent. Worth investigating, I should think.