Consider the involutive transformation $$\mathbb{R}^3 \ni (a,b,c) \overset{\phi}{\mapsto} \left( \frac{a + 2 c}{\sqrt{3}}, \frac{a^2 + a c + c^2}{3} - b , \frac{a - c}{\sqrt{3}}\right)$$
Show that if $P(x) = x^3 + 3 a x^2 + 3 b x + c^3$ has real distinct roots $x_1$, $x_2$, $x_3$ then
$$\tilde P(x) = x^3 + 3 \frac{a+ 2 c}{\sqrt{3}} x^2 + 3 \left( \frac{a^2 + a c + c^2}{3} - b\right) x + \left(\frac{a-c}{\sqrt{3}}\right)^3$$
has a unique real root $x_1'$, and
$$\sqrt[3]{x_1'} =\frac{ \sqrt[3]{x_1} + \sqrt[3]{x_2} + \sqrt[3]{x_3} }{\sqrt{3}} $$
Notes:
One can check that the discriminant under the transformation $\phi$ changes like $\Delta(\tilde{P}) = - \Delta(P)$, hence $\tilde P$ does have a unique real root ( and two non-real conjugate).
We use $\sqrt{3}$ to get an involution, not only a bijection. Loosely we have $(\sqrt[3]{x'_j}) = \mathcal{F}(\sqrt[3] x_j)$, where $\mathcal{F}$ is the normalized Fourier transform on triplets.
To get all of the roots of $\tilde P$ one can proceed as follows: choose determinations for $\sqrt[3]{x_j}$ with product $c$. Then apply the Fourier transform and raise to the cube ( nice thing is that the order of the components is not relevant). This will work even for polynomials with complex coefficients. One can see from here why we have the formula for the discriminant.
Formulas with sum of cubic roots equals another cubic root goes back at least to Ramanujan. Here we have an involution, involving familiar things
One could do a similar involution for cubic forms not necessarily monic-- this is reducible to the monic case.
Posted as a reference.
Any feedback would be welcome!
$\bf{Added:}$ Here is calculation with WA that shows the behavior of the discriminant.
$$\tilde P(x) = x^3 + 3 \frac{a+ 2 c}{\sqrt{3}} x^2 + 3 \left( \frac{a^2 + a c + c^2}{3} - b\right)^2 x + \left(\frac{a-c}{\sqrt{3}}\right)^3$$
– Will Jagy May 08 '23 at 01:48