Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
 A: From your computations, it's enough to show that $n^3+2n=n(n^2+2)\equiv 0\pmod{3}$. If $n\equiv 0\pmod{3}$, you are done, else $n\equiv 1,2\pmod{3}$, in which case $n^2+2\equiv 0\pmod{3}$. 
Reducing it to modular arithmetic may save you a lot of messy multiplication.
A: You want to show that $$(n-1)^3+n^3+(n+1)^3 \equiv 0 \pmod 9.$$ There are $9$ cases to check,


*

*$(0-1)^3+0^3+(0+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$

*$(1-1)^3+1^3+(1+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$

*$(2-1)^3+2^3+(2+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$

*$(3-1)^3+3^3+(3+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$

*$(4-1)^3+4^3+(4+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$

*$(5-1)^3+5^3+(5+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$

*$(6-1)^3+6^3+(6+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$

*$(7-1)^3+7^3+(7+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$

*$(8-1)^3+8^3+(8+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$

A: Since $(n-1)^3 + n^3 + (n+1)^3 = 3n(n^2+2)$ we know that it is divisible by $3$. If we show that $n(n^2+2)$ is also divisible by $3$ then $(n-1)^3 + n^3 + (n+1)^3$ is divisible by $3^2 = 9$. There are three cases to check:


*

*$0(0^2+2) \equiv 0 \cdot 2 \equiv 0 \pmod 3$

*$1(1^2+2) \equiv 1 \cdot 0 \equiv 0 \pmod 3$

*$2(2^2+2) \equiv 2 \cdot 0 \equiv 0 \pmod 3$

A: HINT $\rm\:\ \ mod\ 9:\ (x+3)^3\equiv\:x^3\ \Rightarrow\ f(n+1)\: \equiv\ f(n)\ $ so, by induction, $\rm\ f(n)\equiv\: f(1)\equiv\: 0 $ 
E.g. a specific inductive step: $\rm\: mod\ 9\::\ \ 3^3\:\equiv\: 0^3\ \Rightarrow\ 1^3+ 2^3+3^3\ \equiv\ 0^3+1^3+2^3\ \equiv\ 0\:.$
NOTE $\ $ The period $3$ repeating behavior is illustrated vividly in quanta's calculations, where one can explicitly see the repeating $3$-cycle  $\ -1+0+1,\ \ 0+1-1,\ \ 1-1+0,\ \cdots$
A: To say that $3n^3+6n$ is divisible by $9$ is equivalent to saying that $n^3+2n=n(n^2+2)$ is divisible by $3$. Now split to three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$ (explain why this is true.) Now you have to check just 3 cases.
A: Let $\wp(n) = (n-1)^3  + n^3 + (n+1)^3$. Observe that $\wp(1) = 9$, and so is divisible by $9$. To complete this inductive proof, we want to prove that $\wp(n+1) - \wp(n)$ is divisible by $9$.
$\wp(n+1) - \wp(n) = n^3 + (n+1)^3 + (n+2)^3 - (n-1)^3 - n^3 - (n+1)^3 $
$= (n+2)^3 - (n-1)^3 = 9n^2 + 9n + 9 = 9(n^2 + n +1)$
and so is also divisible by $9$. We have thus shown that $\wp(1) = 0 \mod 9$, and that $\wp(n+1) - \wp(n) = 0 \mod 9$, and so $\wp(2) = \wp(1) + \wp(2) - \wp(1) = 0 \mod 9$, and inductively proceed from there...
A: HINT $\rm\quad\displaystyle  (n+1)^3 + n^3 + (n-1)^3\ =\ 3\:(n+1)\:n\:(n-1) + 9\:n\ =\ 18\:{n+1 \choose 3} - 9\:n$
A: $9|3n^3 + 6n$ , is equivalent to , $3|n^3 + 2n$ , is equivalent to $3|n^3 - n + 3n$ , is equivalent to ,
$3|n^3 - n$ ; now $n^3 - n = n(n-1)(n+1)$ is the product of three consecutive  integers whenever $n$ 
is an integer and since consecutive multiples of $3$ occur with a difference(gap) of $3$ , 
$n(n-1)(n+1)=n^3 - n$ must be divisible by $3$ whenever $n$ is an integer.
