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
 A: The difference from one step to the next is $(n+3)^3-n^3$, which has just a few terms.
A: Hint $\ (n\!-\!1)^3\! + n^3\! + (n\!+\!1)^3\! -9n = 3\underbrace{(n\!-\!1)n(n\!+\!1)}_{\large \rm divisible\ by\ 3}\,$ since they've same roots and lead coef.
A: 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:
$9k + 9n^2+27n+27 = 9(k+n^2+3n+3) = \lambda9$
$Q.E.D$
A: Using induction: Check that $n^3 + (n+1)^3 + (n+2)^3$ is a multiple of 9 when n = 0. 
Next you know that $n^3 + (n+1)^3 + (n+2)^3$ is a multiple of 9 and want to show that $(n+1)^3 + (n+2)^3 + (n+3)^3$ is a multiple of 9. Well, what's the difference between these two sums? 
The bad thing is that it's a lot of time for me to calculate the difference. The good thing is that it's your problem, so don't be lazy and calculate the difference. The solution actually ends up very simple. 
A: Why not let MMA do the proof?
In[159]:= f[m_] := Sum[k^3, {k, m, m + 2}]

In[161]:= Simplify[Mod[f[m], 9], Element[m, Integers]]

Out[161]= 0

Regards,
Wolfgang
A: 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 $3n(n^2 +2)$.
A: Here's an approach using the same basic thinking as modular arithmetic, without actually using modular arithmetic. Any three consecutive numbers will feature one number which  is a multiple of $3$ (call it $3a$), one number $1$ greater than a multiple of $3$ (call it $3b+1$), and one number which is $1$ smaller than a multiple of $3$ (call it $3c-1$). Depending on the particular value of the first of the consecutive numbers, one or more of $a,b,c$ may be the same, but we don't need to rely on that here. We just sum the cubes of the three numbers.
$(3a)^3+(3b+1)^3+(3c-1)^3=27a^3+27b^3+27b^2+9b+1+27c^3-27c^2+9c-1$. 
Note that on the RHS, the $1$ and $-1$ cancel, so the remaining terms reduce to:
$9(3a^3+3b^3+3b^2+b+3c^3-3c^2+c)$, which has an explicit factor of $9$. QED
