Socratis T.n.p.
2020-12-19 03:02:04 UTC
n° Teorema T.n.p. : Somma di 4 cubi.
===>a^3+ (b^3+c^3) + f^3 == 2(b+c)*[(b*c)+(2d)^2)] <=============
d=1) 2^3+ (3^3+4^3) +5^3 == 2(3+4) * [(3*4)+ (2^2)] = 14*16 = 224m^3
d=2) 2^3+ (4^3+6^3) +8^3 == 2(4+6) * [(4 * 6) ++16] = 20*40 = 800m^3
d=3) 2^3+ (5^3+8^3) +11^3==2(5+8) * [(5 * 8) ++36] = 26*76 =1976m^3
d=4) 2^3+ (6^3+10^3)+14^3= 2(6+10)*[(6 *10)++64] =32*124=3968m^3
Saluti da Socratis T.n.p.
Roma.19.dic.2020
===>a^3+ (b^3+c^3) + f^3 == 2(b+c)*[(b*c)+(2d)^2)] <=============
d=1) 2^3+ (3^3+4^3) +5^3 == 2(3+4) * [(3*4)+ (2^2)] = 14*16 = 224m^3
d=2) 2^3+ (4^3+6^3) +8^3 == 2(4+6) * [(4 * 6) ++16] = 20*40 = 800m^3
d=3) 2^3+ (5^3+8^3) +11^3==2(5+8) * [(5 * 8) ++36] = 26*76 =1976m^3
d=4) 2^3+ (6^3+10^3)+14^3= 2(6+10)*[(6 *10)++64] =32*124=3968m^3
Saluti da Socratis T.n.p.
Roma.19.dic.2020