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 $x=\sqrt2+\sqrt[3]5$ for simplicity. By manually expanding we get
\begin{array}{rrrrrrr}
1 = & 1 \\
x = & & 1\cdot 2^{1/2} & + 1\cdot 5^{1/3} \\
x^2 = & 2 & & & + 2\cdot 2^{1/2}5^{1/3} & +1\cdot 5^{2/3} \\
x^3 = & 5 & + 2\cdot 2^{1/2} & + 6\cdot 5^{1/3} & & & +3\cdot2^{1/2}5^{2/3} \\
x^4 = & 4 & + 20\cdot 2^{1/2} & + 5\cdot 5^{1/3} & + 8\cdot2^{1/2}5^{1/3} & +12\cdot5^{2/3} \\
x^5 = & 100 & + 4\cdot 2^{1/2} & + 20\cdot 5^{1/3} & + 25\cdot2^{1/2}5^{1/3} & +5\cdot5^{2/3} & +20\cdot 2^{1/2}5^{1/3} \\
\end{array}
We want to solve $\sqrt2 = a + bx + cx^2 + dx^3 + ex^4 + fx^5$, and this can be rewritten as the matrix equation:
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 0 & 0 & 0\\
2 & 0 & 0 & 2 & 1 & 0 \\
5 & 2 & 6 & 0 & 0 & 3 \\
4 & 20 & 5 & 8 & 12 & 0 \\
100 & 4 & 20 & 25 & 5 & 20
\end{bmatrix}^{\mathrm T}
\begin{bmatrix}a \\ b \\ c \\ d \\ e \\ f\end{bmatrix}
=
\begin{bmatrix}0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{bmatrix},
$$
from which we can find that
$$ -1820 + 735x - 780x^2 - 320x^3 + 45x^4 + 48x^5 = 1187 \sqrt2. $$
$\sqrt[3]5$ is then simply $x-\sqrt2$.
