$\mathbf{Q}[\sqrt 5+\sqrt[3] 2]=\mathbf{Q}[\sqrt 5,\sqrt[3] 2]$?

We don't need any sophisticated tools like Galois Theory here. It is possible to prove this using simple algebraic manipulation. It should be obvious that $\mathbb{Q}(\sqrt{5} + \sqrt[3]{2}) \subseteq \mathbb{Q}(\sqrt{5},\sqrt[3]{2})$ and proving that $\mathbb{Q}(\sqrt{5} + \sqrt[3]{2}) \supseteq \mathbb{Q}(\sqrt{5},\sqrt[3]{2})$ is not difficult. Let $c = \sqrt{5} + \sqrt[3]{2}$ and we need to establish that both $a = \sqrt{5}$ and $b = \sqrt[3]{2}$ are rational functions of $c$.

We have $$(c - a)^{3} = 2\Rightarrow c^{3} - 3ac^{2} + 3a^{2}c - a^{3} = 2$$ or $$c^{3} - 3ac^{2} + 15c - 5a = 2\Rightarrow a = \frac{c^{3} + 15c - 2}{3c^{2} + 5}\tag{1}$$ so that $a$ is a rational function of $c$. And further we have $a + b = c$ so that $b = c - a$ and therefore $b$ is also a rational function of $c$.

Here is an approach using a bit of Galois theory.

Let $K \subset L$ a Galois extension and subfields $K \subset K_1, K_2 \subset L$. We have $K_1 \supset K_2$ if and only if the following hold: whenever $g$ is in the Galois group of $L/K$ such that $g$ fixes any element of $K_1$, $g$ also fixes any element of $K_2$ ( therefore $K_1 \supset K_2$ if and only if $\text{Gal}(L/K_1) \subset \text{Gal}(L/K_2))$.

Consider the same extension $K\subset L$ and $\alpha_1$, $\alpha_2$ in $L$. From the above we conclude: $K(\alpha_1) \supset K(\alpha_2)$ if and only if: for every $g$ in $\text{Gal}(L/K)$ that fixes $\alpha_1$, $g$ also fixes $\alpha_2$.

Consider again $K\subset L$ and $u$, $v$, $w$ $\ldots$ elements in $L$. Let $R(u,v,w,\ldots)$ a rational expression in $u$, $v$, $w$ $\ldots$. Say we want to show that $K ( R(u,v,w,\ldots) )$ contains $u$. Let $u_1=u$, $u_2$ $\ldots$ in $L$ the conjugates of $u$,$\ \ $ $v_1= v$,$v_2$, $\dots$ the conjugates of $v$,$ \ \ $ $w_1 = w$, $w_2$ $\ldots$ ... of $w$. ( in fact it's enough to assume that $u_1=u$, $u_2$, $\ldots$ roots in $L$ of a polynomial with coefficients in $K$ and same for $(v_j)$, $(w_k)$ )

Assume that $$R(u,v,w,\ldots) \ne R(u_i, v_j, w_k,\ldots )$$ if $i \ne 1$. Then $$K ( R(u,v,w,\ldots) )\ni u$$

Indeed, let $g$ in $\text{Gal}(L/K)$. We have $g u = u_i$, $g v = v_j$, $g w = w_k$, $\ldots$ for some indexes $i$, $j$, $k$, $\ldots$ since $g u$ is a conjugate of $u$, $g v$ is a conjugate of $v$ and so on. Therefore we have $$ g( R(u,v,w,\ldots) ) = R(u_i, v_j, w_k, \ldots)$$

Assume that $g( R(u,v,w,\ldots) )= R(u,v,w,\ldots) $. Therefore we have $R(u_i, v_j, w_k, \ldots)= R(u,v,w,\ldots)$. From the condition above we conclude that $i=1$ and therefore $g u = u_1 = u$.

We have showed that $g( R(u,v,w,\ldots) )= R(u,v,w,\ldots) $ implies $g u = u$. Therefore $K ( R(u,v,w,\ldots) )\ni u$.

Application: Let $a_1$, $a_2$ $a_3$, $\ldots$ rational numbers $>0$,$\ $ $e_1$, $e_2$, $\ldots$ natural numbers $>0$. Then we have

\begin{eqnarray} \mathbb{Q}( \sum \sqrt[e_s]{a_s}) = \mathbb{Q}( \sqrt[e_1]{a_1}, \sqrt[e_2]{a_2}, \ldots) \end{eqnarray}

Indeed, $\sum \sqrt[e_s]{a_s}> \text{Real part}( \text{sum of any other set of conjugates} ) $ and so no other set of conjugates has the sum equal to $\sum \sqrt[e_s]{a_s}$ and so \begin{eqnarray} \mathbb{Q}( \sum \sqrt[e_s]{a_s}) \supset \mathbb{Q}( \sqrt[e_{s_0}]{a_{s_0}}) \end{eqnarray} for all $s_0$.

Similarly we can also prove: \begin{eqnarray} \mathbb{Q}( \sqrt[3]{2} \cdot \sqrt{5}) = \mathbb{Q}( \sqrt[3]{2} , \sqrt{5}) \end{eqnarray}

Obs: the nonequality of certain expression involving algebraic numbers can be checked in $\mathbb{C}$ using "real methods" ( inequalities ).

Example: if $\sqrt[3]{2} \cdot \sqrt{5} = \zeta^{j-1} \sqrt[3]{2} \cdot (-1)^k \sqrt{5}$ then $\zeta^{j-1}$ must be real and so $j=1$.

Apologies if this duplicates content in the nice answer of orangeskid. Here is another approach, also using Galois theory.

Let $K$ be the splitting field of the polynomial $(x^{2}-5)(x^{3}-2)$ over $\mathbb{Q}$, and let $\alpha = \sqrt[3]{2}+\sqrt{5}$. One can see that $K = \mathbb{Q}(\sqrt{5}, \sqrt[3]{2}, \xi)$, where $\xi$ is a primitive third root of unity, and hence is the normal closure of the subfield $L := \mathbb{Q}(\sqrt{5}, \sqrt[3]{2})$. Now, $[L:\mathbb{Q}] = 6$, since $\deg(\sqrt{5}) = 2$ and $\deg(\sqrt[3]{2}) = 3$ both divide $[L:\mathbb{Q}]$, and $\mathbb{Q}(\sqrt[3]{2}+\sqrt{5})$ is clearly a subfield of $L$. If we can show that $\deg(\alpha) = 6$, the conclusion follows.

Since $L \subset \mathbb{R}$ and $\xi \notin \mathbb{R}$, it follows that $[K:L] = 2$, whence $[K:\mathbb{Q}] = 12$. Then $G := \mathrm{Gal}(K/\mathbb{Q})$ has $12$ elements, and any element $\sigma \in G$ is determined by what it does to the generators $\sqrt{5}, \sqrt[3]{2}, \xi$ of $K$ over $\mathbb{Q}$. Any $\sigma \in G$ must satisfy $\sigma(\sqrt{5}) \in \{\pm \sqrt{5}\}, \sigma(\sqrt[3]{2}) \in \{\xi^{i}\sqrt[3]{2}\}_{i = 0}^{2}, \sigma(\xi) \in \{\xi, \xi^{2}\}$, so counting, there are at most 12 possible automorphisms, arising from these possible assignments on generators. Since we have already determined $|G| = 12$, every assignment on generators described above must give rise to an automorphism of $K$ over $\mathbb{Q}$. Hence, let $\sigma, \tau \in G$ be the automorphisms defined on generators by $$\sigma(\sqrt[3]{2}) = \xi\sqrt[3]{2}, \sigma(\sqrt{5}) = \sqrt{5}, \sigma(\xi) = \xi$$ $$\tau(\sqrt[3]{2}) = \sqrt[3]{2}, \tau(\sqrt{5}) = -\sqrt{5}, \tau(\xi) = \xi$$

Then the Galois orbit $G\alpha$ contains $\{\pm \sqrt{5} + \xi^{i}\sqrt[3]{2} \mid i = 0, 1, 2\}$, so $|G\alpha| \geqslant 6$. Since $|G\alpha| \leqslant 6$, we obtain the equality $|G\alpha| = 6$. Further, as the minimal polynomial of $\alpha$ over $\mathbb{Q}$ factors as $\prod_{s \in G\alpha} (X-s)$, it follows that $\deg(\alpha) = 6$, as desired.