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 $9$
 A: 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.
A: Need to show that
$$x^3+3x^2+5x+3 =0 \mod 3$$
That is
$$x(x^2+5)=0 \mod 3$$
It's true when $x=0 \mod 3$.
For $$x=\pm 1 \mod 3$$,
$$(3k + 1)((3k + 1)^2+5)=(3k + 1)(9k^2+6k+6)=0 \mod 3$$,
$$(3k - 1)((3k - 1)^2+5)=(3k - 1)(9k^2-6k+6)=0 \mod 3$$.
A: $(x-1)^3+x^3+(x+1)^3=3x^3+6x=3(x^3+2x)=3x(x^2+2)$. 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 $x^2+2=9k^2+12+4+2=3(3k^2+4k+2)$.
