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?

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.
