Proof $(n^2 + 1)$ in $\Theta(n^3 - 2n - 3)$ I trying resolve this problem, but the first equation have 3 terms and the second equation have 2 terms.
I don't know how to resolve this problem. Any idea?

Note that
$\Theta(n^3 - 2n - 3)$ 
is the set of functions $f(n)$
such that
$\frac{f(n)}{n^3 - 2n - 3}$
is bounded away from both $0$ and $\infty$
for arbitrarily large $n$.
Since $\frac{n^2+1}{n^3 - 2n - 3} \to 0$
as $n \to \infty$,
$n^2+1 \notin \Theta(n^3 - 2n - 3)$.
However,
$n^2+1 \in O(n^3 - 2n - 3)$.
This is often written
$n^2+1 = O(n^3 - 2n - 3)$.
Also note that
$\Theta(n^3 - 2n - 3) = \Theta(n^3)$. 
