Hi I have this so far but im a little unsure on how should I write to continue, a ny hint would be greatful since I'm learning recently:
Prove, using induction, that the sum of the cubes of three consecutive natural numbers is divisible by nine, that is, that $(n-2)^3 + (n-1)^3 + n^3 = 9 $ , (x) for some $x \in \mathbb{N} $.
Base case:( n = 2 )
$(2-2)^{3} + (2-1)^{3} +2^{3} = 0^{3} + 1^{3} + 2^{3} = 0+1+ 8 = 9$
Which is true because it is divisible by 9.
P.I: Sup. $(n-2)^{3}+(n-1)^{3}+n^{3} = 9x $ so with n+1\ $((n+1)-2)^{3}+((n+1)-1)^{3}+(n+1)^{3} = 9x $
I think it's supposed to continue with the $(3+2)-2 ^{3}$ etc... but after?