Let numbers are $a, a+1,a+2$ $\implies a^3+(a+1)^3+(a+2)^3=3a^3+9a^2+15a+9$ Then: $3a+3|3a^3+9a^2+15a+9$ given that $(3a+3).(a^2+2a+3)=3a^3+9a^2+15a+9$ And finished

parmenides51 wrote: Show the the sum of any three consecutive positive integers is a divisor of the sum of their cubes. Let numbers are $a, a+1,a+2$ then: $a^3+(a+1)^3+(a+2)^3=3a^3+9a^2+15a+9$ Then: $3a+3|3a^3+9a^2+15a+9$ then: $(3a+3).(a^2+2a+3)=3a^3+9a^2+15a+9$

Albert123 wrote: Let numbers are $a, a+1,a+2$ $\implies a^3+(a+1)^3+(a+2)^3=3a^3+9a^2+15a+9$ Then: $3a+3|3a^3+9a^2+15a+9$ given that $(3a+3).(a^2+2a+3)=3a^3+9a^2+15a+9$ And finished e61442289 wrote: parmenides51 wrote: Show the the sum of any three consecutive positive integers is a divisor of the sum of their cubes. Let numbers are $a, a+1,a+2$ then: $a^3+(a+1)^3+(a+2)^3=3a^3+9a^2+15a+9$ Then: $3a+3|3a^3+9a^2+15a+9$ then: $(3a+3).(a^2+2a+3)=3a^3+9a^2+15a+9$ your solutions seems identical, e61442289 you had better avoid such posts in order to avoid being accused of copying other's solutions