Proving symmetric polynomial identity using generating functions

Question

Let be $e_{k}(X_{n})=e_{k}(x_{1},x_{2}, \dots, x_{n})$ the $k$-th elementary symmetric polynomial on the variables $x_{1},x_{2}, \dots, x_{n}$.

I managed to prove the following proposition using manipulations of symmetric polynomials identities:

Let be $f_{a,n},f_{b,n}, f_{c,n}$ defined as the following $$f_{a,n}=\sum_{k=0}^{n-2}(-1)^{k}\frac{(n-k-1)(n-k)}{2}x_{1}^{n-k-2}e_{k}(X_{n})$$ $$f_{b,n}=\sum_{k=1}^{n-1}(-1)^{k}k(n-k)x_1^{n-k-1}e_{k}(X_{n})$$ $$f_{c,n}=\sum_{k=2}^{n}(-1)^{k}\frac{k(k-1)}{2}x_1^{n-k}e_{k}(X_{n})$$ Then $$f_{b,n}^2-4f_{a,n}f_{c,n}=(n-1)^2\displaystyle \prod_{i=2}^{n}(x_{1}-x_{i})^2$$

Is it possible to get a generating function proof for it? Any hint?

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When $x_1$ is replaced for $x_m$, $1 \leq m \leq n$, the argument on the product of the right side changes to $(x_m-x_i)^2$ for $i\neq m$. Therefore,

Let be $f_{a,n},f_{b,n}, f_{c,n}$ defined as the following $$f_{a,n}=\sum_{k=0}^{n-2}(-1)^{k}\frac{(n-k-1)(n-k)}{2}x_{m}^{n-k-2}e_{k}(X_{n})$$ $$f_{b,n}=\sum_{k=1}^{n-1}(-1)^{k}k(n-k)x_m^{n-k-1}e_{k}(X_{n})$$ $$f_{c,n}=\sum_{k=2}^{n}(-1)^{k}\frac{k(k-1)}{2}x_m^{n-k}e_{k}(X_{n})$$ Then $$f_{b,n}^2-4f_{a,n}f_{c,n}=(n-1)^2\displaystyle \prod_{i\neq m}^{}(x_{m}-x_{i})^2$$

A Sage code to explore the formula is available here

var_list=["x"+str(i) for i in range(0,100)]+["e"+str(i) for i in range(0,100)]
var(var_list+["n"])
e = SymmetricFunctions(QQ).e()
def ele_dict(deg):
        es=[e[i].expand(deg) for i in range(0,deg+1)]
        coe=list([eval("e"+str(i)) for i in range(0,deg+1)])
        return dict(zip(coe,es))
def asum(n):
    return sum((-1)**k * (n-k-1)*(n-k)/2 * x0**(n-k-2) *eval("e"+str(k)) for k in [0..(n-2)])
def bsum(n):
    return sum((-1)**k * k*(n-k) * x0**(n-k-1) *eval("e"+str(k)) for k in [1..(n-1)])
def csum(n):
    return sum((-1)**k * k*(k-1)/2 * x0**(n-k) *eval("e"+str(k)) for k in [2..(n)])
for j in range(3,9):
    formula=bsum(j)**2-4*asum(j)*csum(j)
    che=formula.subs(ele_dict(j)).full_simplify().factor()
    print(che)