Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded 
Let $n \in \mathbb N$ and:
  $$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
  Prove $x_n$ is bounded sequence.

Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right)
\end{align}
$$
From here:
$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 - {1\over n^2}} \lt 1
$$
Therefore:
$$
\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0
$$
Which means $x_n \gt 0$.
Consider the following inequality:
$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\ 
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2
$$
Also $x_n \gt0$ so finally:
$$
0 < x_n <2
$$
Have i done it the right way?
 A: $$x_n=\frac{(n^3+1)^2-(n^2-1)^3}{\sqrt[3]{(n^3+1)^6}+\sqrt[3]{(n^3+1)^5}\sqrt{n^2-1}+...}\rightarrow0,$$
which says that $\{x_n\}$ is bounded. 
A: This may be weaker than the other ways (that are nice, including OP's), but seemed like an easy way to see it; note that $n \le \sqrt[3]{n^3+1} \le n+1$, and $n-1 \le \sqrt{n^2-1} \le n$ [can be verified by cubing & squaring, respectively]
So $\sqrt[3]{n^3+1} - \sqrt{n^2-1} \ge n - n = 0$ and $\sqrt[3]{n^3+1} - \sqrt{n^2-1} \le (n+1) - (n-1) = 2$.
A: It seems right and it is bounded, as a minor observation we can improve the upper bound
$$x_n < \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} \le\sqrt 2$$
A: Your prove is fine but a lot more work than necessary.
As $n \ge 1$ we have
$n = \sqrt[3]{n^3} < \sqrt[3]{n^3 + 1} < \sqrt[3]{n^3 + 3n^2 + 3n + 1} = \sqrt[3]{(n+1)^3} = n+1$
and 
$n = \sqrt{n^2} > \sqrt{n^2 -1 } = \sqrt{n^2 - 2 + 1} \ge \sqrt{n^2 - 2n + 1} = \sqrt{(n-1)^2} = n-1$.
So $0 = n - n < \sqrt[3]{n^3 + 1} - \sqrt{n^2 -1} < (n+1) - (n-1) = 2$.
A: $$\sqrt[3]{n^3+1}-\sqrt{n^2-1}=\sqrt[3]{n^3+1}-n-\sqrt{n^2-1}+n
\\=\frac1{(n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2}+\frac1{\sqrt{n^2-1}+n}$$
and both terms are decreasing. The supremum is with $n=1$,
$$x_1=\sqrt[3]2.$$
