Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!


  Problem:
  
  For any natural number $n , n^3 + 2n$ is divisible by $3.$
  This makes sense
  
  Proof:
  
  Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
  $2 \times 0 = 0.$ So it is divisible by $3.$
  
  Induction: Assume that for an arbitrary natural number $n$,
  $n^3+ 2n$ is divisible by $3.$ 
  
  Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
  the induction hypothesis. Got it
  
  $$( n + 1 )^3+ 2( n + 1 ) = (  n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
  
  $$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
 and regrouping}\}$$
  $$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
  
  which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
  by the induction hypothesis. What?


Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
In the inductive hypothesis, you assumed that $n^3 + 2n$ was divisible by 3 for some $n$, and now you're proving the same for $n+1$. It's like knocking down dominoes: if you can prove that the first domino falls over (base case) and each domino knocks over the next (inductive step), then that means that all of the dominoes get knocked down eventually.

You know that $(n^3 + 2n) + 3(n^2 + n + 1)$ is divisible by 3 because $n^3 + 2n$ is (because of the inductive hypothesis) and $3(n^2 + n + 1)$ is (because it's 3 times an integer). So, their sum is as well.
RTP : $a^3+3a$  is in the form of $3k$

Proof : $a^3+2a=
            a^3-a+3a=
            a(a^2-1)+3a=
            a(a+1)(a-1)+3a$
We know that sum of three consecutive numbers is divisible by 3           $$3k+3a=3(k+a)=3k$$