This question can be answered by general theory, at least when $\Delta$ is a fundamental discriminant (so it's not a square times a smaller discriminant). Assuming this, the general procedure for finding the forms of discriminant $\Delta$ that represent a number $n>1$ primitively is the following. Let the prime factorization of $n$ be $n=p_1^{e_1}\cdots p_k^{e_k}$ for distinct primes $p_i$. If $n$ is represented primitively by some form of discriminant $\Delta$ then each $p_i$ will also be represented (primitively). Let $Q_i$ be a form representing $p_i$. This is unique up to equivalence, giving an element in the narrow class group $CG(\Delta)$ that I will also call $Q_i$. Thus in $CG(\Delta)$ the two elements representing $p_i$ are $Q_i$ and $Q_i^{-1}$ (which coincide when $Q_i$ is symmetric, or "ambiguous" in Gauss's terminology). The statement is then that the elements of $CG(\Delta)$ representing $n$ primitively are precisely the products $Q_1^{\pm e_1}\cdots Q_k^{\pm e_k}$, with the restriction that $e_i$ must be $1$ for each $p_i$ dividing $\Delta$. The $2^k$ choices of signs for the exponents give $2^k$ potentially different elements of $CG(\Delta)$ representing $n$ primitively, though often the number is smaller due to relations that hold in $CG(\Delta)$ and to some forms $Q_i$ being symmetric.
For representing negative numbers $n$ when $\Delta>0$ one applies the preceding process to find the forms representing $|n|$ and then one takes the negatives of these forms to get the forms representing $n$.
Thus for the question of finding the forms representing $-a^3$ primitively one first applies the preceding remarks to find all the forms representing $a$ primitively, then one takes the cubes of these forms in $CG(\Delta)$ to get all the forms representing $a^3$ primitively, and finally one takes the negatives of these forms to get the forms representing $-a^3$ primitively. The condition that $e_i=1$ for primes $p_i$ dividing $\Delta$ means that $a$ must be relatively prime to $\Delta$ in order to have $-a^3$ represented primitively.
As an example consider discriminant $\Delta =136=8\cdot 17$. Here the class group is cyclic of order $4$ generated by the nonsymmetric form $Q=3x^2+2xy-11y^2$. In $CG(\Delta)$ we have $Q^2=34x^2-y^2$, $Q^3=Q^{-1}=3x^2-2xy-11y^2$, and $Q^4=x^2-34y^2$, the principal form giving the identity element of $CG(\Delta)$. It is not hard to see that $Q$ is equivalent to $-Q$, and even properly equivalent to it, say by drawing a small portion of its Conway topograph which has a 180 degree rotational "skew-symmetry" interchanging its positive and negative values. Thus if we take $a=3$, then $3$ is represented by $Q$ since $Q(1,0)=3$ and hence, passing to $CG(\Delta)$, $27$ is represented primitively only by $Q^{\pm3}=Q^{\pm1}$, hence $-27$ is represented primitively only by $-Q^{\pm3}=Q^{\pm1}$. All these forms are equivalent to $Q$. Explicitly we have $Q(7,4)=27$ and $Q(4,3)=-27$.
If we take $a=9$, this is represented primitively only by $Q^2=34x^2-y^2$, when $(x,y)=(1,5)$, so $9^3$ is represented primitively only by $Q^6=Q^2$ and $-9^3$ is represented primitively only by $-Q^2=x^2-34y^2$.
If we take $a=15=3\cdot 5$, then since $3$ and $5$ are both represented by $Q$, $15$ is represented primitively only by $Q^{\pm1}Q^{\pm1}$ with independent choices of signs, so $15$ is represented primitively only by $Q^0=x^2-34y^2$ (when $(x,y)=(7,1)$) and by $Q^2=34x^2-y^2$ (when $(x,y)=(2,11)$). So $15^3$ is represented primitively only by $Q^0$ and $Q^6=Q^2$, and $-15^3$ only by their negatives $Q^2$ and $Q^0$.
It is worth noting that $Q$ is equivalent to its negative but $x^2-34y^2$ is not equivalent to its negative $-x^2+34y^2$.
When $\Delta$ is not a fundamental discriminant, say $\Delta=d^2\Delta'$ with $\Delta'$ a fundamental discriminant, there are extra complications, although if one restricts attention to representing numbers relatively prime to $d$ then the theory described above still works.