<s>[INST] Proving $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ I'm trying to prove by MI. I have already distributed n+1, but now I'm stuck on how I can show 9 divides the RHS since $42n$ and $3n^3$ does not divide evenly. 
$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36$$
 [/INST] If the property is true for $n$, then it is true for $n+9$ (all three cubes will increase by a multiple of 9). So it suffices to check for $n=0,1,...8$.
$$0^3+1^3+2^3=9=1\cdot 9$$
$$1^3+2^3+3^3=36=4\cdot 9$$
$$2^3+3^3+4^3=99=11\cdot 9$$
$$3^3+4^3+5^3=216=24\cdot 9$$
$$4^3+5^3+6^3=405=45\cdot 9$$
$$5^3+6^3+7^3=684=76\cdot 9$$
$$6^3+7^3+8^3=1071=119\cdot 9$$
$$7^3+8^3+9^3=1584=176\cdot 9$$
$$8^3+9^3+10^3=2241=249\cdot 9$$
Actually, as you showed that the sum is a multiple of $3$, it even suffices to show that the average is a multiple of $3$ for $n=0, 1, 2$.
$$\frac{0^3+1^3+2^3}3=3=1\cdot 3$$
$$\frac{1^3+2^3+3^3}3=12=4\cdot 3$$
$$\frac{2^3+3^3+4^3}3=33=11\cdot 3$$
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