Solution:

Let $S(n)=n^3+2 n$ is divisible by 3 .

Step 1:

Inductive base: for $n=1$

$(1)^3+2.1=3$ which is divisible by 3

Thus, $s(1)$ is true.

step2 :

Inductive hypothesis: Let $s(k)$ is true ide. $k^3+2 k$ is divisible by 3 holds true.

$ \text { or, } k^3+2 k=3 s \text { for } s \in N $

Step 3::-

Inductive step: We have to show that $s(k+1)$ is true.

Consider, $(k+1)^3+2(k+1)$ is divisible by 3 .

Now, $(k+1)^3+2(k+1)=k^3+1+3 k^2+3 k+2 k+2$\

$ =\left(k^3+2 k\right)+3\left(k^2+k+1\right)\\ $

$=3 s+3 l\\$ where, $l=k^2+k+1 \in N\$

$=3(s+l)\\$ Therefore, $s(k+1)$ is true. Hence, by principle of mathematical induction $s(n)$ is true for all $n \in N$.