Răspuns :
(3k-1)³+(3k)³+(3k+1)³=
=27k³-27k²+9k-1+27k³+27k³+27k²+9k-1=
=27k³+9k+27k³+27k³+9k=
=81k³+18k=
=9k(9k²+2) deci divizibil cu 9
=27k³-27k²+9k-1+27k³+27k³+27k²+9k-1=
=27k³+9k+27k³+27k³+9k=
=81k³+18k=
=9k(9k²+2) deci divizibil cu 9