The sum of three consecutive cubes numbers produces 9 multiple I want to prove that $n^3 + (n+1)^3 + (n+2)^3$ is always a $9$ multiple
I used induction by the way.
I reach this expression:  $(n+1)^3 + (n+2)^3 + (n+3)^3$
But is a lot of time to calculate each three terms, so could you help me to achieve the induction formula 
Thanks in advance

There is one really elegant and simple way of proving this by means of induction. 
It goes as following: 
Define the problem as: $n^3 + (n+1)^3 + (n+2)^3 = 9\lambda$ 
Test for $n=1$; $1^3+2^3+3^3=9\lambda$; $36=4*9$ 
Assume that for some $n \in N$ -> $n^3 + (n+1)^3 + (n+2)^3 = 9\lambda$  

Test for $n+1$ 
$
 (n+1)^3+(n+2)^3+(n+3)^3 = 9\lambda
$ 
$
(n+1)^3+(n+2)^3+(n+3)(n^2+6n+9)=9\lambda
$$(n+1)^3+(n+2)^3+n^3+9n^2+27n+27$
According to our assumption $n^3+(n+1)^3+(n+2)^3$ is divisible by 9 so we can put
$9k$ instead giving us:
$9k + 9n^2+27n+27 = 9(k+n^2+3n+3) = \lambda9$
$Q.E.D$
