<s>[INST] Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. I have the solution for this but I get lost at the end, here's what I have so far. 
basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$ 
      $9 \mid  1 + 8$ = true
Induction: Assume $n^3 + (n + 1)^3 + (n + 2)^3 = k * 9$ // Why set it equal to $k * 9$? I know it works but why not just make the assumption => $n^3 + (n + 1)^3 + (n + 2)^3$ for some $n = k \ge 0$ 
Then, //and here's where I get lost
$(n + 1)^3 + (n + 2) + (n + 3)^3 = k * 9 + [(n + 3)^3 - n^3] = 9 (k + n^2 + 3n + 3)$  
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I've done similar examples but none like this. What am I not seeing?  
 [/INST] You assume $\;9\mid\left[n^3+(n+1)^3+(n+2)^3\right]\;$ , and now you want to prove that also
$$9\mid\left[(n+1)^3+(n+2)^3+(n+3)^3\right]\;\;,\;\;\text{but}$$
$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36=$$
$$=\left(3n^3+9n^2+15n+9\right)+9n^2+27n+27$$
and you can see the left part between the parentheses is divisible by nine by the inductive hypothesis, whereas the second part is obviously divisible by nine, too...
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