Discriminant of $f(x)=x^3+ax+b$ Suppose we have the polynomial $f(x)=x^3+ax+b$, with roots $\alpha, \beta, \gamma$ in $\mathbb{C}$, and let $\Delta = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$.
Is there any quick way of show that the discriminant $\Delta^2$ is $-4a^3-27b^2$?
It seems as though we need to exploit certain properties of symmetric functions. Is there some way of writing out $\Delta$ in terms of elementary symmetric polynomials perhaps?
 A: $\Delta$ is not expressible in terms of elementary symmetric polynomials (because it is not invariant by all permutations of the roots, and so $-4a^3-27b^2$ is not a square).
The standard way is to develop $\Delta^2$ and write that in terms of elementary symmetric polynomials.
I guess you can take a shortcut by noticing that $\Delta^2$ is homogeneous of degree $6$ in the roots, so it has to be a linear combination of $(\alpha\beta\gamma)^2$ and $(\alpha\beta+\alpha\gamma+\beta\gamma)^3$ (the other products involving $\alpha+\beta+\gamma$ vanish because this is $0$ in our case)
So you only have to figure out the two coefficients. Then by picking some custom polynomials for which you know the roots, like $X(X-1)(X+1)$ and $(X-1)(X-2)(X+3)$, the discriminant is simple to compute, $a$ and $b$ are too, so you should be able to deduce the right coefficients from there.
A: As the others have said, you need to consider $Δ^2$. Then by sorting the factors you can assemble them to form the product (assuming for the moment that none of the roots is zero, i.e., $b\ne 0$)
\begin{align}
-Δ^2&=Res_x(f(x),f'(x))
=f'(α)f'(β)f'(γ)
=(3α^2+a)(3β^2+a)(3γ^2+a)\\[.5em]
&=-(2a+\tfrac{3b}α)(2a+\tfrac{3b}β)(2a+\tfrac{3b}γ)
=\frac1b(2aα+3b)(2aβ+3b)(2aγ+3b)
\end{align}
which you can then combine using
$$
0=(α+β+γ)^2=α^2+β^2+γ^2+2a
$$

One can see the initial identity in terms of formal derivatives of polynomials or ordinary derivatives of polynomial functions with complex coefficients. If extending the argument to a splitting field for a polynomial over a general domain, the formal derivative only can be used.
\begin{align}
f'(x)&=3x^2+a
=\frac{d}{dx}(x-α)(x-β)(x-γ)\\
&=(x-α)(x-β)+(x-β)(x-γ)+(x-α)(x-γ).
\end{align}
Then 
$$
f'(α)=3α^2+a=(α-β)(α-γ)
$$ 
etc. so that the product of the three derivatives, after three sign changes, is $Δ^2$,
\begin{align}
f'(α)f'(β)f'(γ)
&=(α-β)(α-γ)\,(β-α)(β-γ)\,(γ-α)(γ-β)\\
&=(-1)^3\,(α-β)^2(β-γ)^2(γ-α)^2
\end{align}
