For $n =3,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}, e_{3}.]$(manually) Here is the question I want to answer letter $(e)$ of it manually:
Let $B = \mathbb{Q}[x_{1}, ... , x_{n}] \cong \mathbb{Q}^{[n]}$ and $A = \mathbb{Q}[e_{1}, ... , e_{n}]$ where $e_{i} \in B$ is the elementary symmetric polynomial of degree $i.$ The discriminant is the polynomial  $\Delta^2,$ where $\Delta \in B$ is defined by:\
$$\Delta = \prod_{i > j}(x_i -x_j)$$
$(a)$ (Vandermonde determinant) Show that: $$ \Delta = \det
\begin{pmatrix}
1 & x_{1} & x_{1}^2 & \dots & x_{1}^{n-1}\\
1 & x_{2} & x_{2}^2 & \dots & x_{2}^{n-1}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & x_{n} & x_{n}^2 & \dots & x_{n}^{n-1}
\end{pmatrix}$$
$(b)$  Show that $\Delta^2 \in B^{S_{n}}.$ Deduce that  $\Delta^2 \in A.$
$(c)$ Show that $\Delta \notin B^{S_{n}}.$ Determine which subgroup of $S_{n}$ fixes $\Delta.$
$(d)$ For $n =2,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}].$
$(e)$  For $n =3,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}, e_{3}].$
Here is the solution using Mathematica:



Here is the hint given to me to solve it manually:



Still I can not see how this hint will make the solution easier manually, could anyone help me in that please?
 A: What a strange hint. There's a much better way to do it, and I have no idea why the author didn't set $a_3 = 1$ from the start.
I learned this argument from David Speyer. Computing the discriminant as a polynomial in the $e_i$ is equivalent to computing the discriminant of the polynomial $f(x) = x^3 - e_1 x^2 + e_2 x - e_3$. Up to a translation, which does not affect the discriminant, we can remove the quadratic term from this polynomial: if we substitute $x = y + \frac{e_1}{3}$, then we get a new polynomial (brace yourself)
$$\begin{align} g(y) &= \left( y^3 + e_1 y^2 + \frac{e_1^2}{3} y + \frac{e_1^3}{27} \right) - e_1 \left( y^2 + \frac{2e_1}{3} y + \frac{e_1^2}{9} \right) + e_2 \left( y + \frac{e_1}{3} \right) - e_3 \\
 &= y^3 + \left( - \frac{e_1^2}{3} + e_2 \right) y + \left( -\frac{2e_1^3}{27} + \frac{e_1 e_2}{3} - e_3 \right). \end{align}$$
Set $p = - \frac{e_1^2}{3} + e_2, q = -\frac{2e_1^3}{27} + \frac{e_1 e_2}{3} - e_3$, so that this new polynomial can be written $g(y) = y^3 + py + q$. Now it suffices to compute the discriminant of this new polynomial as a function of $p$ and $q$, then plug in these expressions for $p$ and $q$ in terms of the $e_i$.
The point of doing this is that we've now removed most of the terms from the discriminant: $p$ has degree $2$, $q$ has degree $3$, and $\Delta^2$ has degree $6$, which implies that it must be some linear combination $ap^3 + bq^2$ of $p^3$ and $q^2$. So we only have two coefficients to compute. We can actually compute these coefficients by computing the discriminant of two specific polynomials:

*

*Set $p = -1, q = 0$. Then $g(y) = y^3 - y = y(y + 1)(y - 1)$ has roots $0, \pm 1$, so its discriminant is $\left( (0 - 1)(1 - (-1))( (-1) - 0) \right)^2 = 4$, which gives $a = -4$.

*Set $p = 0, q = -1$. Then $g(y) = y^3 - 1 = (y - 1)(y - \omega)(y - \omega^2)$ has roots $1, \omega, \omega^2$ where $\omega$ is a primitive third root of unity (satisfying $1 + \omega + \omega^2 = 0$), so its discriminant is $\left( (1 - \omega)(\omega - \omega^2)(\omega^2 - 1) \right)^2 = (1 - \omega)^6 = (- 3 \omega)^3 = -27$, which gives $b = -27$.

So we get $\boxed{ \Delta^2 = -4p^3 - 27q^2 }$ (already a useful formula in practice), and substituting $e_1, e_2, e_3$ in gives the full cubic discriminant (of more dubious value; I have never used it).
