Question

represent 3 cube as sum of three consecutive numbers​

Answer 1

So here's my proof:

Let a ∈ Z+

Define

S(x)=x³+(x+1)³+(x+2)³

So,

S(a)=a³+(a+1)³+(a+2)³

S(a)=a³+(a³+3a²+3a+1)+(a³+6a²+12a+8)

S(a)=3a³+9a²+15a+9

S(a)=3(a³+3a²+5a+3)

Hence, 3 ∣S(a).