Show that $n^3+2n$ is divisible by 3 for all $n\ge 1$ i want to prove it with mathematical induction :
first i am tried with n=0
then it is divisible by zero then i move to next step change all n with K then i get this product :
$$P(K)=K^3+2K = 3m$$
Note: $3k$ because we multiply any no. with $3$ is divisible by $3$
now , next step i am increase $+1$ in $k$ so i get this step :
$$(K+1)^3 + 2(K+1) = 3m$$
so now next step i am not able to solve please help.

I know you want to prove by induction, but anothter proof would be to observe that every integer is congruent to either 0, 1, or 2 (mod 3). Then we want to see if for every integer n:
$n^3 + 2n \equiv 0\pmod{3}$
Well, take all integers n such that $n \equiv 0\pmod{3}$. Then $n^3 \equiv 0\pmod{3}$ and $2n \equiv 0\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
Take all integers n such that $n \equiv 1\pmod{3}$. Then $n^3 \equiv 1\pmod{3}$ and $2n \equiv 2\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
Take all integers n such that $n \equiv 2\pmod{3}$. Then $n^3 \equiv 2\pmod{3}$ and $2n \equiv 4\equiv1\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
