New algorithm for cube (third power) computation

[Alberto Zanoni]

> Maybe it's easier to do something useful for fifth power? E.g., using
> that
>
>   (x^3 + 10) (x^2 + 5x + 10) = x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 50 x + 100
>
> "masters"
>
>   (x + 1)^5 = x^5 + 5 x^4 + 10 x^3 + 10 x^2 + 5x + 1
>
> ?

One should not forget the a_i and b_i coefficients: it seems that in this case 
the problem is the final recomposition of longer coefficients than in the 
cube case.

I tryed to do some experiments with fifth power, but for what I have analyzed 
it seems that recomposing (longer) coefficients a0^5, a1a0^5 etc. mixes 
things up such in a way that the cube trick does not work any more (or at 
least I was not able to see anything useful).

Anyway, I'll try again to have a look as soon as possible.
-- 
Alberto Zanoni
Centro Interdipartimentale "Vito Volterra"
Universita' degli Studi di Roma "Tor Vergata"
Via Columbia 2
00133 Roma, Italia
http://bodrato.it/papers/zanoni.html