Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$

Let $u = \sqrt{2} + \sqrt[3]{5}$, we have
$$\begin{align}
(u - \sqrt{2})^3 = 5
\iff & u^3 - 3\sqrt{2}u^2 + 6u - 2\sqrt{2} = 5\\
\implies &
\begin{cases}
\sqrt{2} &= \frac{u^3 + 6u - 5}{3u^2 + 2} \in \mathbb{Q}(u)\\
\sqrt[3]{5} &= u - \sqrt{2} \in \mathbb{Q}(u)
\end{cases}
\end{align}
$$
As a result, $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) \subset \mathbb{Q}(u)$.
The other direction is trivial because
$$\sqrt{2}, \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})
\implies u = \sqrt{2} + \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})
\implies \mathbb{Q}(u) \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$
