Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Teorema : (n-d) * n * (n+d) = n^3 - n*d^2

17 views
Skip to first unread message

Socratis T.n.p.

unread,
Jan 8, 2022, 2:00:24 PM1/8/22
to
Teorema : (n-d) * n * (n+d) = n^3 - n*d^2

5 * 6 * 7 = 6^3 -(6*1^2) = 210m^3
5 * 7 * 9 = 7^3 -(7*2^2) = 315m^3

10 * 13 * 16 = 13^3 -(13*3^3) = 2'080m^3
10i *20i *30i = i20^3 -(i20*i10^2)=6000i^3

10i *20i *30i === 6000i^3
1 * 2 * 3 = 6m^3=6000i^3
20i * 30i = 600i^2 = 6m^2

Felice anno 2022 ==> 4.5*9*18 = 9^3.

Socratis T.n.p.

unread,
Jan 8, 2022, 5:57:05 PM1/8/22
to
Il giorno sabato 8 gennaio 2022 alle 20:00:24 UTC+1 Socratis T.n.p. ha scritto:
> Teorema : (n-d) * n * (n+d) = n^3 - n*d^2
>
> 5 * 6 * 7 = 6^3 -(6*1^2) = 210m^3
> 5 * 7 * 9 = 7^3 -(7*2^2) = 315m^3

Teorema : 0.5n * n*2n = n^3
========> 3.5 *7 *14 = 7^3
0.5 *1 * 2 = 1000i^3.....Alias :
5i *10i *20i = i10^3 =1000i^3 =1m^3. Socratis.
0 new messages