Use induction to prove that $n^3 + (n+1)^3 + (n+2)^3 $ is divisible by $9$ Prove that for all integers $n\geq 0, n^3 + (n+1)^3 + (n+2)^3 $ is divisible by 9.


*

*If $ n=1, 1+8+27 = 36 = 9 * x $

*Suppose $ n = k, k^3 + (k+1)^3 + (k+2)^3 $ is divisible by 9.

*Find out $ n = k + 1, $ is divisible by 9.
Because $2$ is divisible by $9$. if $3$ is divisible by $9$, $3$ - $2$ will be divisible by $9$
and the result of $3$ - $2$ =  $ 9(k^2 + 3k + 3) $ which is divisible by $9$. So, it is proved.
What I found out the answer so far is them. Are they right way?
 A: The write up is rather confused; it is particularly bad to use "2" and "3-2" as you do, since it seems you are saying that the number $2$ is divisible by $9$, that $3-2$ (that is, that $1$) is divisible by $9$,  etc.
Spend the time writing out complete, coherent, self-contained sentences! Confused writing usually indicates confused thinking.
More specific comments:


*

*Bad use of $x$ in your first step. Better to write out explicitly that $36$ is equal to $9\times 4$, hence divisible by $9$.

*Write out explicitly what your assumption is: namely, your assumption is that there exists an integer $q$ such that
$$ k^3 + (k+1)^3 + (k+2)^3 = 9q.$$

*Write out explicitly what you need to prove. Namely, you need to prove that
$$(k+1)^3 + (k+2)^3 + (k+3)^3\text{ is divisible by }9.$$
Then you make a very bad mistake: you affirm the consequent. You are trying to prove that the expression in point 3 is divisible by $9$. You assume that it is divisible by 9, and then point out that if it is divisible by 9, then so is 
$$\Bigl( (k+1)^3 + (k+2)^3 + (k+3)^3\Bigr) - \Bigl( k^3 + (k+1)^2 + (k+2)^3\Bigr).$$
Then you note that this is indeed divisible by $9$, and conclude that the assumption must be true.
This is a logical fallacy. You are saying: "If $P$, then $Q$; since $Q$ is true, then $P$ must be true." If I fall into a pool, I'll get wet. I'm wet. Therefore, I fell into a pool. (Well, no, there are other reasons why I may be wet).
Don't assume what you are trying to prove. Deduce it.
Instead, what you want to do is to write out the expression you want, and try to use the fact that $k^3 + (k+1)^3 + (k+2)^3$ is divisible by $9$ to conclude the expression you want is divisible by $3$. For instance, you can write:
$$\begin{align*}
(k+1)^3 + (k+2)^3 + (k+3)^3 &= (k+1)^3 + (k+2)^3 + \Bigl( k^3 + 9k^2 + 27k + 81\Bigr)\\
&= k^3 + (k+1)^3 + (k+2)^3 + \Bigl( 9k^2 + 27k + 81\Bigr)
\end{align*}$$
and go from there to conclude that if $k^3 + (k+1)^3 + (k+2)^3$ is divisible by $9$, then $(k+1)^3 + (k+2)^3 + (k+3)^3$ is also divisible by $9$.
This completes the Inductive Step. Together with the Base (step 1), this establishes the desired result for all integers.
A: Hint $\rm\ \ \  \bbox[1px,border:3px solid #c60]{\bbox[8px,border:1px solid #c00]{\forall\, n\ge 0\!:\ 9\ |\ f_n\!\iff\! [\,9\ |\ f_0\,\ {\rm  and}\  \rm\:\forall\, n\ge 0\!:\ 9\ |\ \color{#0a0}{f_{\,n+1}\!-f_n}\,]}}\ \ $ (typical telescopy)
because, $ $ if $\rm\ \ 9\ |\ \color{#0a0}{f_{n+1}\!-f_n}\ $ then $\rm\ 9\ |\ \color{#c00}{f_n} \Rightarrow\ 9\ |\ f_{n+1} =  \color{#0a0}{f_{n+1}\! - f_n} + \color{#c00}{f_n}\ \,$  (inductive step $\,\rm P_n\Rightarrow P_{n+1}$)
i.e. $\rm\: mod\ 9\!:\ if\ \ \color{#0a0}{f_{n+1}\equiv f_n}\ $ then $\rm\ \color{#c00}{f_n\equiv c}\ \,\Longrightarrow\,\ \color{#0a0}{f_{n+1}\equiv \color{#0a0}{f}_{\color{#c00} n}}\color{#c00}{\equiv c}\ $ so induction yields
that $\rm\:f\:$ is a constant function, so $\rm\ f_n\equiv f_0\equiv\: 0^3+1^3+2^3\equiv 9\equiv 0$.
This holds in OP by $\rm\ f_{n+1}-f_n = (n\!+\!3)^3\!-n^3 = 9\ (n^2\! +\! 3n\! +\! 3)\equiv 0\pmod{\!9}$.
This is a special case of telescopic induction. For further examples see many prior posts.
