Show that $\mathbb{Q}(\sqrt{5}+\sqrt[3]{2})=\mathbb{Q}(\sqrt{5},\sqrt[3]{2})$.

Since you know that $[\mathbb Q[\sqrt[3]{2},\sqrt 5]:\mathbb Q] = 6$, you know that each of the six values $$1,\sqrt{5},\\\sqrt[3]{2},\sqrt[3]{2}\sqrt{5},\\\sqrt[3]{4},\sqrt[3]{4}\sqrt 5\tag{1}$$ are linearly independent over $\mathbb Q$.

Now $$(\sqrt[3]2+\sqrt 5)^2=5\cdot 1 + 2\cdot \sqrt[3]2\sqrt 5 + 1\cdot\sqrt[3]4$$

Is it possible for $1,\sqrt[3]2+\sqrt5,(\sqrt[3]2+\sqrt 5)^2$ to be linearly dependent over $\mathbb Q$?

Do the same with by adding the cube $(\sqrt[3]2+\sqrt5)^3$.

Another way to look at it use (1) as a basis, and write elements of the field as:

$$(a,b,c,d,e,f)\to a\cdot 1 + b\cdot \sqrt5+c\sqrt[3]2+d\sqrt[3]2\sqrt5+e\sqrt[3]4+f\sqrt[3]4\sqrt5$$

Then $$\begin{align}(1,0,0,0,0,0)&\leftrightarrow 1\\(0,1,1,0,0,0)&\leftrightarrow \sqrt 5+ \sqrt[3]2\\(5,0,0,2,1,0)&\leftrightarrow (\sqrt5+\sqrt[3]2)^2\\ (2,5,15,0,0,3)&\leftrightarrow (\sqrt5+\sqrt[3]2)^3 \end{align}$$

And those four vectors are "obviously" linearly independent.

Let's show that $s=\sum \sqrt[n_i]{d_i}$ ($d_i >0$ rationals ) generates $\mathbb{Q}(\sqrt[n_i]{d_i})$. Consider all in larger Galois extension $K \supset \mathbb{Q}(\sqrt[n_i]{d_i})\supset \mathbb{Q}$. Now, to show that $s$ generates all the $\sqrt[n_i]{d_i}$ it's enough to show that whenever a Galois transformation $\phi$ of $K$ preserves $s$ it must preserve all the $\sqrt[n_i]{d_i}$. Now any Galois transformation $\phi$ takes $\sqrt[n_i]{d_i}$ to some $\omega_i \sqrt[n_i]{d_i}$ where $\omega_i^{n_i} =1$, so $\phi(s) = \sum \omega_i \sqrt[n_i]{d_i}$. Note that $$\left|\sum_i \omega_i \sqrt[n_i]{d_i}\right| \le \sum_i \sqrt[n_i]{d_i}$$ with equality if and only if $\omega_i$ have the same argument. Therefore, if $\sum_i \omega_i \sqrt[n_i]{d_i}= \sum_i \sqrt[n_i]{d_i}$ then all the $\omega_i$ must be $1$.

Conclusion: $\phi(s) = s \implies \phi( \sqrt[n_i]{d_i}) = \sqrt[n_i]{d_i}$ for all $i$.

$\bf{Added:}$ A solution that uses mostly linear algebra, inspired from @ Thomas Andrews: 's solution.

We have the equalities \begin{eqnarray} \left(\begin{array} {c} 1\\\sqrt[3]{2}+\sqrt{5}\\ (\sqrt[3]{2}+\sqrt{5})^2\\ (\sqrt[3]{2}+\sqrt{5})^3\\ (\sqrt[3]{2}+\sqrt{5})^4\\ (\sqrt[3]{2}+\sqrt{5})^5 \end{array}\right) = \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 5 & 0 & 1 & 0 & 2 & 0 \\ 2 & 15 & 0 & 5 & 0 & 3 \\ 25 & 2 & 30 & 8 & 20 & 0 \\ 100 & 125 & 2 & 25 & 10 & 50 \\ \end{array} \right) \cdot \left(\begin{array} {c} 1 \\ \sqrt[3]{2}\\ \sqrt[3]{4} \\ \sqrt{5} \\ \sqrt[3]{2} \sqrt{5} \\\sqrt[3]{4} \sqrt{5} \end{array} \right) \end{eqnarray} hence \begin{eqnarray} \left(\begin{array} {c} 1 \\ \sqrt[3]{2}\\ \sqrt[3]{4} \\ \sqrt{5} \\ \sqrt[3]{2} \sqrt{5} \\\sqrt[3]{4} \sqrt{5} \end{array} \right) = \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ \frac{2275}{4054} & -\frac{1714}{2027} & \frac{195}{2027} & \frac{500}{2027} & -\frac{9}{4054} & -\frac{30}{2027} \\ \frac{2875}{2027} & -\frac{1325}{2027} & -\frac{955}{2027} & \frac{150}{2027} & \frac{100}{2027} & -\frac{9}{2027} \\ -\frac{2275}{4054} & \frac{3741}{2027} & -\frac{195}{2027} & -\frac{500}{2027} & \frac{9}{4054} & \frac{30}{2027} \\ -\frac{6505}{2027} & \frac{1325}{4054} & \frac{1491}{2027} & -\frac{75}{2027} & -\frac{50}{2027} & \frac{9}{4054} \\ -\frac{5143}{2027} & \frac{2335}{2027} & -\frac{650}{2027} & -\frac{991}{2027} & \frac{15}{2027} & \frac{100}{2027} \\ \end{array} \right) \cdot \left(\begin{array} {c} 1\\\sqrt[3]{2}+\sqrt{5}\\ (\sqrt[3]{2}+\sqrt{5})^2\\ (\sqrt[3]{2}+\sqrt{5})^3\\ (\sqrt[3]{2}+\sqrt{5})^4\\ (\sqrt[3]{2}+\sqrt{5})^5 \end{array}\right) \end{eqnarray}

For instance, we have

$ \tiny \sqrt{5} = -\frac{2275}{4054} + \frac{3741}{2027} (\sqrt[3]{2}+\sqrt{5}) -\frac{195}{2027}(\sqrt[3]{2}+\sqrt{5})^2 -\frac{500}{2027} (\sqrt[3]{2}+\sqrt{5})^3+ \frac{9}{4054} (\sqrt[3]{2}+\sqrt{5})^4+ \frac{30}{2027} (\sqrt[3]{2}+\sqrt{5})^5$

Brute-Force Method

Let $\alpha:=\sqrt{5}+\sqrt[3]{2}$. Then, $\alpha^3-3\sqrt{5}\alpha^2+15\alpha-5\sqrt{5}=(\alpha-\sqrt{5})^3=2$, whence $$\left(\alpha^3+15\alpha-2\right)^3=5\left(3\alpha^2+5\right)^2\,,$$ or $\alpha$ is a root of the polynomial $$f(x):=x^6-15x^4-4x^3+75x^2-60x-121 \in \mathbb{Q}[x]\,.$$ If this polynomial is reducible, then consider it modulo $3$, so $$f(x)=x^6+2x^3+2=\left(x^2+2x+2\right)^3$$ in $\mathbb{F}_3$, and $x^2+2x+2$ is an irreducible element of $\mathbb{F}_3[x]$. That is, $f(x)$ must have a monic quadratic factor $g(x)$ in $\mathbb{Q}[x]$, and by Gauss's Lemma, $g(x)\in\mathbb{Z}[x]$. Hence, $g(x)=x^2+ax+b$ with $a,b\in\mathbb{Z}$ and $b\equiv 2 \equiv -1\pmod{3}$. Let $f(x)=g(x)\,h(x)$ for some $h(x)\in\mathbb{Q}[x]$ (which again yields $h(x)\in\mathbb{Z}[x]$). Since the coefficient of $x^5$ in $f(x)$ is $0$, $h(x)=x^4-ax^3+cx^2+dx+e$ for some $c,d,e\in\mathbb{Z}$.

Now, $$f(x)=x^6-4x^3+4=\left(x^3-2\right)^2=\left(x^3+2^3\right)^2=(x+2)^2\left(x^2-2x+4\right)^2$$ in $\mathbb{F}_5$, where both $x+2$ and $x^2-2x+4$ are irreducible elements of $\mathbb{F}_5[x]$. Hence, in $\mathbb{F}_5$, either $$g(x)=(x+2)^2=x^2+4x+4\text{ or }g(x)=x^2-2x+4\,.$$ In either case, $b\equiv 4\equiv -1\pmod{5}$. Together with $b\equiv -1\pmod{3}$, we deduce that $b \equiv -1\pmod{15}$. However, $b$ must divide the constant term $-121=-11^2$ of $f(x)$. This means $b=-1$ or $b=-121$.

If $b=-1$, then $g(x)=x^2+ax-1$ and $h(x)=x^4-ax^3+cx^2+dx+121$. Equating the coefficients of $f(x)$ and $g(x)\,h(x)$, we have $$c-a^2-1=-15\,,\,\, ac+a+d=-4\,,$$ $$ad-c+121=75\,,\text{ and }121a-d=-60\,.$$ Consequently, $c=a^2-14$ and $d=121a+60$, so that $$\begin{align}0&=(ad-c+121)-75=ad-c+46\\&=a(121a+60)-\left(a^2-14\right)+46=60\left(2a^2+a+1\right)\,.\end{align}$$ However, $2a^2+a+1=0$ does not have an integer solution.

If $b=-121$, $g(x)=x^2+ax-121$ and $h(x)=x^4-ax^3+cx^2+dx+1$. Equating the coefficients of $f(x)$ and $g(x)\,h(x)$, we have $$c-a^2-121=-15\,,\,\, ac+121a+d=-4\,,$$ $$ad-121c+1=75\,,\text{ and }a-121d=-60\,.$$ Consequently, $c=a^2+106$ and $d=\frac{a+60}{121}$, so that $$\begin{align}0&=(ad-121c+1)-75=ad-121c-74\\&=\frac{a}{121}(a+60)-121\left(a^2+106\right)-75\,,\end{align}$$ or $$0=-\frac{1}{121}\left(14640a^2-60a+1561021\right)\,.$$ However, $14640a^2-60a+1561021=0$ does not have an integer solution.

We have derived a contradiction from the hypothesis that $f(x)$ is reducible over $\mathbb{Q}$, so $f(x)$ must be irreducible in $\mathbb{Q}[x]$. Hence, $\mathbb{Q}(\alpha)\cong \mathbb{Q}[x]/\big(f(x)\big)$ is a field extension of index $\deg(f)=6$ over $\mathbb{Q}$.