For any positive integer $n$, let $S(n)$ denote the statement
$$
S(n) : 6\mid (n^3+11n).
$$

**Base step:** For $n=1, S(1)$ gives $1^3+11(1) = 12 = 2\cdot 6$. Thus, $S(1)$ holds.

**Inductive step:** Let $k\geq 1$ be fixed, and suppose that $S(k)$ holds; in particular, let $\ell$ be an integer with $6\ell = k^3+11k$. Then
\begin{align}
(k+1)^3 + 11(k+1) &= (k^3+3k^2+3k+1) + (11k+11)\tag{expand}\\[0.5em]
&= \color{red}{k^3+11k}+3k^2+3k+12\tag{rearrange}\\[0.5em]
&= \color{red}{6\ell} + 3k(k+1)+2\cdot 6.\tag{by ind. hyp.}
\end{align}
Since one of $k$ and $k+1$ is even, the term $3k(k+1)$ is divisible by $6$, and so the last expression above is divisible by $6$. This proves $S(k+1)$ and concludes the inductive step $S(k)\to S(k+1)$.

By mathematical induction, for each $n\geq 1$, the statement $S(n)$ is true. $\blacksquare$