What this is all about: Aronhold's invariants S, T are polynomial functions of the coefficients of a ternary cubic form which are invariant under the action of SL₂(R). They generate the graded algebra of invariants, so two ternary cubic forms are equivalent over the reals if and only if the corresponding ordered pairs of invariants (S,T) match. It's hard to find formulas for S and T written down on the web, so I typed them up here:

Ternary cubic form: a*x^3 + b*y^3 + c*z^3 + 3*d*x^2*y + 3*e*y^2*z + 3*f*z^2*x + 3*g*x*y^2 + 3*h*y*z^2 + 3*i*z*x^2 + 6*j*x*y*z

S = a*g*e*c - a*g*h^2 - a*j*b*c + a*j*e*h + a*f*b*h - a*f*e^2 - d^2*e*c + d^2*h^2 + d*i*b*c - d*i*e*h + d*g*j*c - d*g*f*h - 2*d*j^2*h + 3*d*j*f*e - d*f^2*b - i^2*b*h + i^2*e^2 - i*g^2*c + 3*i*g*j*h - i*g*f*e - 2*i*j^2*e + i*j*f*b + g^2*f^2 - 2*g*j^2*f + j^4

T = a^2*b^2*c^2 - 3*a^2*e^2*h^2 - 6*a^2*b*e*h*c + 4*a^2*b*h^3 + 4*a^2*e^3*c - 6*a*d*g*b*c^2 + 18*a*d*g*e*h*c - 12*a*d*g*h^3 + 12*a*d*j*b*h*c - 24*a*d*j*e^2*c + 12*a*d*j*e*h^2 - 12*a*d*f*b*h^2 + 6*a*d*f*b*e*c + 6*a*d*f*e^2*h + 6*a*i*g*b*h*c - 12*a*i*g*e^2*c + 6*a*i*g*e*h^2 + 12*a*i*j*b*e*c + 12*a*i*j*e^2*h - 6*a*i*f*b^2*c + 18*a*i*f*b*e*h - 24*a*g^2*j*h*c - 24*a*i*j*b*h^2 - 12*a*i*f*e^3 + 4*a*g^3*c^2 - 12*a*g^2*f*e*c + 24*a*g^2*f*h^2 + 36*a*g*j^2*e*c + 12*a*g*j^2*h^2 + 12*a*g*j*f*b*c - 60*a*g*j*f*e*h - 12*a*g*f^2*b*h + 24*a*g*f^2*e^2 - 20*a*j^3*b*c - 12*a*j^3*e*h + 36*a*j^2*f*b*h + 12*a*j^2*f*e^2 - 24*a*j*f^2*b*e + 4*a*f^3*b^2 + 4*d^3*b*c^2 - 12*d^3*e*h*c + 8*d^3*h^3 + 24*d^2*i*e^2*c - 12*d^2*i*e*h^2 + 12*d^2*g*j*h*c + 6*d^2*g*f*e*c - 24*d^2*j^2*h^2 - 12*d^2*i*b*h*c - 3*d^2*g^2*c^2 - 24*g^2*j^2*f^2 + 24*g*j^4*f - 12*d^2*g*f*h^2 + 12*d^2*j^2*e*c - 24*d^2*j*f*b*c - 27*d^2*f^2*e^2 + 36*d^2*j*f*e*h + 24*d^2*f^2*b*h + 24*d*i^2*b*h^2 - 12*d*i^2*b*e*c - 12*d*i^2*e^2*h + 6*d*i*g^2*h*c - 60*d*i*g*j*e*c + 36*d*i*g*j*h^2 + 18*d*i*g*f*b*c - 6*d*i*g*f*e*h + 36*d*i*j^2*b*c - 12*d*i*j^2*e*h - 60*d*i*j*f*b*h + 36*d*i*j*f*e^2 + 6*d*i*f^2*b*e + 12*d*g^2*j*f*c - 12*d*g*j^3*c - 12*d*g*j^2*f*h + 36*d*g*j*f^2*e - 12*d*g*f^3*b + 24*d*j^4*h + 12*d*j^2*f^2*b + 4*i^3*b^2*c + 24*i^2*g^2*e*c - 27*i^2*g^2*h^2 - 36*d*j^3*f*e - 12*i^3*b*e*h + 8*i^3*e^3 - 24*i^2*g*j*b*c + 36*i^2*g*j*e*h + 6*i^2*g*f*b*h + 12*i^2*j^2*b*h - 3*i^2*f^2*b^2 - 12*d*g^2*f^2*h - 12*i^2*g*f*e^2 - 24*i^2*j^2*e^2 + 12*i^2*j*f*b*e - 12*i*g^3*f*c + 12*i*g^2*j^2*c + 36*i*g^2*j*f*h - 12*i*g^2*f^2*e - 36*i*g*j^3*h - 12*i*g*j^2*f*e + 12*i*g*j*f^2*b + 24*i*j^4*e - 12*i*j^3*f*b + 8*g^3*f^3 - 8*j^6

Example: If our ternary form is x^3+y^3+z^3, then the corresponding ordered pair (S,T) is (0,1). Furthermore, the class number of the cubic form x₁^3 + ... + x_n^3 is 1 for any n (exercise for the reader), so in particular any integral ternary cubic form with invariants (S,T) = (0,1) is equivalent to x^3+y^3+z^3 over the integers. Thus, such a form has the property that a, b, and c can each be represented as sums of three (possibly negative) cubes. For instance, the cubic form

3x^3 + 2y^3 + z^3 - 3x^2y - 42y^2z + 12z^2x - 3xy^2 + 66yz^2 + 12zx^2 + 24xyz

has (S,T) = (0,1), and corresponds to the representations 3 = 4^3 + 4^3 - 5^3, 2 = -5^3 - 6^3 + 7^3, 1 = 10^3 + 9^3 - 12^3.

j-invariant: The j-invariant can be expressed in terms of S and T by the formula j = 1728*(4S)^3/((4S)^3 - T^2). This provides a way to calculate the j-invariant of a cubic curve, even if it has no rational points.