# For the $4x^3-g_{2}x-g_{3}=0$, why the discriminant it must be of the form $\alpha g_{2}^3+\beta g_{3}^2$?

For the $4x^3-g_{2}x-g_{3}=0$, why the discriminant it must be of the form $\alpha g_{2}^3+\beta g_{3}^2$? Is there any proof about it?

The discriminant is the square of the product of the differences of the roots, so it is a homogeneous symmetric polynomial of degree $6$. The Fundamental Theorem of Symmetric Polynomials implies that it can be written as a polynomial in $s_i$, the elementary symmetric polynomials in the roots (and these are related to the coefficients of the equation; in particular in this case, $s_1=0$, $s_2=g_2/4$, $s_3=-g_3/4$.
Since the discriminant is homogeneous, it must be a linear combination of monomials, all of which must have degree $6$, which means we need terms of the form $s_1^as_2^bs_3^c$ with $a+2b+3c=6$, $a,b,c \geq 0$. The form of the cubic implies that $s_1=0$, so we can forget about any term that contains an $s_1$. Then we need solutions to $2b+3c=6$, $b,c \geq 0$. There are only two, $b=3,c=0$ and $b=0,c=2$. Hence the discriminant is is a linear combination of $s_2^3 \propto g_2^3$ and $s_3^2 \propto g_3^2$.
For a cubic equation of the form $x^3 + ax^2 + bx + c$, the discriminant is calculated to be $a^2b^2 + 18abc − 4b^3 − 4a^3c − 27c^2$. Seeing your equation only has the $b=-g_2/4,c=-g_3/4$ coefficients, the only nonzero parts in the discriminant are $-4b^3 -27c^2$, which is an equation in terms of the cube of $g_2$ and the square of $g_3$.