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,

, 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.

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LaTeX renderer refuses to work properly. Sorry.

By:

Alexandre Borovikon May 4, 2010at 2:18 pm

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

By:

Todd Trimbleon May 4, 2010at 3:36 pm

A nice derivation!

By:

mishaon May 6, 2010at 12:32 am