Posted by: Alexandre Borovik | May 4, 2010
Andreas Caicedo found recently what seems to be the best way of remembering the cubic formula. It makes sense over, say, $\mathbb{C}$, and when expanded gives the usual formula, which makes sense over any field. By translating we may assume a given cubic has the form
$x^3 + ax + b = 0$.
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 $x$ (!)
Indeed  if
$c=\cos (3 \theta)$,
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 $z$ of $latex y$, and
$latex x=\cos (\theta) = (z+1/z)/2$.
Decoding this gives the usual formulas.