Prove by mathematical induction that: $\forall n \in \mathbb{N}: 3^{n} > n^{3}$
There is a flaw in the statement you are trying to prove; it is simply false for when $n=3$, since $$ 3^3 \not> 3^3. $$ What you are looking to establish, I suspect, is that $n^3 < 3^n$ for all $n\geq 4$. We can prove this using induction.
Start by noting that $$ 3n^2+3n+1<2(3^n)\tag{1} $$ is true for $n\geq 4$. One can verify $(1)$ using induction or, more cumbersomely, in a direct fashion.
Claim: For $n\geq 4$, $$ n^3 < 3^n. $$
Proof. For $n\geq 4$, let $P(n)$ denote the proposition $$ P(n) : n^3 < 3^n. $$
Base step ($n=4$): Since $4^3=64<81=3^4$, the statement $P(4)$ is true.
Inductive step: Suppose that for some fixed $k\geq 4$, $$ P(k) : k^3 < 3^k $$ holds. It must be shown that $$ P(k+1) : (k+1)^3 < 3^{k+1} $$ follows. Starting with the left-hand side of $P(k+1)$, \begin{align} (k+1)^3 &= k^3+3k^2+3k+1\\[0.5em] &< 3^k+3k^2+3k+1\tag{by $P(k)$}\\[0.5em] &< 3^k+2(3^k)\tag{by $(1)$}\\[0.5em] &= 3(3^k)\\[0.5em] &= 3^{k+1}, \end{align} we end up with the right-hand side of $P(k+1)$. Thus, $P(k+1)$ is also true, and this concludes the inductive step $P(k)\to P(k+1)$.
Thus, by mathematical induction, $P(n)$ is true for all $n\geq 4$. $\blacksquare$
Hint
- For $n=1,2,3$ it's true:$3^1\geq 1^3$, $3^2\geq 2^3$ and $3^3\geq 3^3$
- for some $n\geq 3$, suppose that :$3^k\geq k^3$ for all $k\leq n$ ,so $3^{n-1}\geq (n-1)^3$ but also $3^2\geq 2^3$ hence : $$3^{n+1}=3^{n-1}.3^2\geq 2^3.(n-1)^3=(2(n-1))^3 $$ and now you only need to show that $2(n-1)\geq n+1$ which is true.