Answer
$f(n)$ is divisible by $3$ for all values of $n$.
Work Step by Step
Suppose that $f(n)=n^3-n$; we need to prove that $f(n)$ is divisible by $3$.
1. Our aim is to find that $3|n^3-n$ is true for $n=1$
$n^3-n=3m \implies m=0$
So, it is true for $n=1$.
2. Our aim is to find that $f(n)$ is true for $n=k$.Thus, it will also true for $n=k+1$
$(k+1)^3-(k+1)=3p \implies (k+1)^3-(k+1)=k^3+3k^2+3k+1-k-1 $
Since, $k^3-k=3m$
This yields: $ (k+1)^3-(k+1)=3m+3k^2+3k \implies (k+1)^3-(k+1)=3(m+k^2+k) $
and $(k+1)^3-(k+1)=3p$
So, it is true for $n=k+1$.
Therefore, $f(n)$ is divisible by $3$ for all values of $n$.
