Andreas Caicedo found recently what seems to be the best way of remembering the cubic formula. It makes sense over, say, , and when expanded gives the usual formula, which makes sense over any field. By translating we may assume a given cubic has the form

.

By scaling, we may further assume it has the form

$latex  x^3 – {\frac{3}{4}} x – {\frac{c}{4}} =0 $.

But this means that

$latex c=\cos (3\theta)$,

where

$latex x=\cos (\theta)$.

Using Euler’s identity, this allows us to find (!)

Indeed  if

,

then

$latex y=e^{{i}\cdot {3\theta}}$

is the root of a quadratic,

$latex y^2 – 2c y  + 1 =0 $.

And $latex e^{i\ theta}$  is the cubic root of $latex y$, and

$latex x=\cos (\theta)  = (z+1/z)/2$.

Decoding this gives the usual formulas.