<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] As an alternative to induction, we take any $3$ consecutive cubes as follows:
$$(n-1)^3 + n^3 + (n+1)^3$$
$$= 3n^3  + 6n$$
$$=3n(n^2 +2)$$
Notice that 
$$\begin{align}n(n^2 + 2) &\equiv n(n^2-1)\pmod 3 \\&\equiv (n-1)(n)(n+1)\pmod 3 \end{align}$$
Since either one of $(n-1),n$ or $(n+1)$ must be divisible by $3$, it follows that$3|n(n^2+2)$. This implies that $3\cdot3=9$ divides $3n(n^2 +2)$.
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