<s>[INST] Proof using deductive reasoning I need to deductively prove that the sum of cubes of $3$ consecutive natural numbers is divisible by $9$. I can prove deductively that they are divisible by $3$ but so far any combination I choose fails to prove the divisibility by $9$. As far as I can see. This is a high school question though, so if someone can explain it to me in a highschool math language, it will be appreciated. Now here is how I try to do it. 
Let $X$ stand for any natural number and let $X+1$ and $X+2$ stand for the two consecutive numbers. I will be cubing, expanding and simplifying them
\begin{align*}
 & (x)^{3}+(x+1)^{3}+(x+2)^{3}\\ 
 &= x^3+x^3+3 x^2+3 x+1+x^3+6 x^2+12 x+8\\ 
 &=3x^{3}+9x^{2}+15x+9 \\ 
 &= 3\left ( x^{3}+3x^{2}+5x+3  \right )\\ 
\end{align*}
This can be used to deductively prove that the sum of cube of $3$ consecutive numbers is divisible by $3$ but I can't prove it is divisible by $9$
 [/INST] $(x-1)^3+x^3+(x+1)^3=3x^3+6x=3(x^3+2x)=3x(x^2+2)$. Now we just have to prove $3|x$ or $3|x^2+2$.
Case $1: x=3k$, then $3|x$.
Case $2: x=3k+1$, then $x^2+2=9k^2+6k+1+2=3(3k^2+2k+1)$.
Case $3: x=3k+2$, then $x^2+2=9k^2+12+4+2=3(3k^2+4k+2)$.
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