Analyse convergence of: $\displaystyle \sum^{\infty}_{n=3} \left (\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right)$

If you are able to prove that both series$$\sum_{n=3}^\infty\sqrt[3]{n^3+2}-n\quad\text{and}\quad\sum_{n=3}^\infty\sqrt{n^2+7}-n\tag1$$converge, or that one of them converges and the other one diverges, then you're done.

Now, note that\begin{align}\sqrt[3]{n^3+2}-n&=\sqrt[3]{n^3+2}-\sqrt[3]{n^3}\\&=\frac2{\sqrt[3]{n^3+2}^2+\sqrt[3]{n^3+2}\sqrt[3]{n^3}+\sqrt[3]{n^3}^2}\\&=\frac2{\sqrt[3]{n^3+2}^2+\sqrt[3]{n^3+2}\,n+n^2}\end{align}and therefore$$\lim_{n\to\infty}\frac{\sqrt[3]{n^3+2}-n}{\frac1{n^2}}=\frac23$$and therefore the first series from $(1)$ converges.

Can you take it from here?

Hint Use that when $\alpha\not = 0$, \begin{equation} (1 + x)^\alpha - 1 \sim \alpha x \end{equation} when $x\longrightarrow 0$.