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
$latex c=\cos (3\theta)$,
$latex x=\cos (\theta)$.
Using Euler’s identity, this allows us to find x (!)
Indeed  if
c=\cos (3 \theta),
$latex y=e^{{i}\cdot {3\theta}}$
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.


  1. LaTeX renderer refuses to work properly. Sorry.

  2. That’s okay — rendered or not, it’s very neat.

  3. A nice derivation!

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