# 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 equation: $(n+1)^3 + (n+2)^3 + (n+3)^3$

But is a lot of time to calculate each three terms, so could you help me to achieve the induction formula

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There are many ways to get to this result, some more efficient than others. At the less efficient end, try expanding the expression rather than simplifying it. $n^3 + (n + 1)^3 + (n + 2)^3$ expands to $3n^3 + 9n^2 + 15n + 9$. Chuck $9n^2 + 9$ to leave us with $3n^3 + 15n$. If $n \equiv 1 \mod 9$, then $3n^3 + 15n \equiv 3 + 6 \mod 9$. If $n \equiv 2 \mod 9$, then $3n^3 + 15n \equiv 6 + 3$. And so on and so forth. – Robert Soupe Jul 8 '14 at 3:28
Not to nitpick, but "equations" have equals signs. I don't see an equation anywhere in your question. – Tim Seguine Jul 8 '14 at 10:49

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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The difference from one step to the next is $(n+3)^3-n^3$, which has just a few terms.

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I calculated the result and I have: $3n^3$+$18n^2$+ $42n$+ $36$. How can I prove the induction method? – Luis Armando Jul 7 '14 at 22:59
For the inductive step, you just need $(n+3)^3-n^3=9n^2+27n+27$. Since the value at $n=0$ is a multiple of 9, and the difference at each stage is a multiple of 9,... – Michael Jul 7 '14 at 23:23

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.

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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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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.

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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

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Interesting, but the OP is trying to determine what the proof is, not convince him/herself that the assertion is true. – Caleb Jul 8 '14 at 14:19