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100^6p+1 ca suma de patru cuburi perfecte rezolvare

Răspuns :

OmuBacovian

[tex]\displaystyle 100^{6p+1}=100^{6p}\cdot 100=100^{6p}\cdot (1+8+27+64)= \\(100^{2p})^3(1^3+2^3+3^3+4^3)=\\\bold{(100^{2p})^3+(100^{2p}\cdot2)^3+(100^{2p}\cdot3)^3+(100^{2p}\cdot4)^3}[/tex]



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