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 $9$

It may be more useful to have the center number be $x$, and the two numbers to either side be $x-1$ and $x+1$.  You have then the sum of three consecutive cubes is $(x-1)^3+x^3+(x+1)^3 = 3x^3+6x=3x(x^2+2)$.
Now, note that either $x$ is a multiple of $3$ or $(x^2+2)$ is a multiple of three.
Proof:  $x=3k\Rightarrow x\equiv 0\pmod{3}$
$x=3k\pm 1\Rightarrow x^2 \equiv (\pm 1)^2 \equiv 1\pmod{3}\Rightarrow x^2+2\equiv 0\pmod{3}$
As $3x(x^2+2)$ will have a multiple of three occurring once in the $3$, and once in either the $x$ or the $(x^2+2)$ term, we have that the sum of three consecutive cubes is a multiple of nine.
