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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.

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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 $\phi(5,1) = -27.$