<s>[INST] 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
 [/INST] 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:
$9k + 9n^2+27n+27 = 9(k+n^2+3n+3) = \lambda9$
$Q.E.D$
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