I believe I have the gist of how to prove this. My professor worked out a problem similar to this one only, instead of ($s_n^3$), he used ($s_n^2$), and I am slightly confused as to how he came up with certain portions of his proof. The following is the proof he gave us for ($s_n^2$). I believe after understanding his proof better, I can prove the original problem more easily. So please do not post the solution to the original question.
Proof {the convergence of the sequence ($s_n$) implies the convergence of ($s_n^2$)}
Since the lim ($s_n$)=s, we know ($s_n$) is bounded.
That is there exists $M\in R$ such that $|s_n|$ $\le$ M for all $n\in N$
Now, for every $\varepsilon >0$ we have lim ($s_n$)$=s$. Working on $\varepsilon/(M+|s|)>0$, there exists $N\in R$ such that $|s_n-s| \le \varepsilon /(M+|s|)$ whenever $n>N$, therefore for all $n>N$, $|s_n^2 - s^2| = |s_n - s|*|s_n + s| \le |s_n - s|(|s_n|+|s|) \le |s_n - s|*(M + |s|)< \varepsilon $
Which proves lim $(s_n^2)$ = $s^2$.
The following is my proof for the current problem (that is in the title).
Let me know if I did anything incorrect.
Proof
Since the lim ($s_n$)=s, we know ($s_n$) is bounded.
That is there exists $M > 0$ such that $|s_n|\le M$ for all $n\in \mathbb{N}$
Now, for every $\varepsilon >0$ since lim ($s_n$)=s, working on $\varepsilon /(3M^2)>0$,
there exists $N\in \mathbb{N}$ such that $|s_n-s| < \varepsilon /3M^2$ whenever $n>\mathbb{N}$
Therefore, for all $n>\mathbb{N}$
$|s_n^3 - s^3|$ = $|s_n - s|$ $|s_n^2 + s_n*s + s^2| \le $ $|s_n - s|$ $(|s_n^2|+|s_n||s|+ |s^2|) \le $ $(|s_n|^2+|s_n||s|+ |s|^2) \le $ $|s_n - s|*(M^2 + M*M + M^2) \le $ $|s_n - s|*(3M^2)< \varepsilon $
Which proves lim $(s_n^3)$ = $s^3$