Răspuns :
cele 3 numere pot fi de forma:
3k,3k+1 ,3k+2
3k+3k+1+3k+2=9k+3=3(3k+1)
==>3(3k+1) etse divizibila cu 3
3k,3k+1 ,3k+2
3k+3k+1+3k+2=9k+3=3(3k+1)
==>3(3k+1) etse divizibila cu 3
Răspuns (a-b)3=a³-3a²b+3ab²-b³ ; ( x-1)³ x³-3x²+3x-1 +x³+3x²+3x+1+x³=3x³+6x=3 (x³+2x) unde numerele sunt x-1,x ,x+1
Explicație pas cu pas: