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 $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$
 A: 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
$$\sqrt{2}, \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})
\implies u = \sqrt{2} + \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2},\sqrt[3]{5})
\implies \mathbb{Q}(u) \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$
A: 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 $x-\sqrt2$.
A: Let $K = \Bbb{Q}(\sqrt{2},\sqrt[3]{5})$ and observe that $N_{K/\Bbb{Q}}(\sqrt{2}+\sqrt[3]{5}) = 17$. Then the following general result implies that $\sqrt{2}+\sqrt[3]{5} \in \mathcal{O}_K$ generates $K$ over $\Bbb{Q}$.

Proposition: Let $K$ be an algebraic number field and $\alpha \in \mathcal{O}_K$. If $N_{K/\Bbb{Q}}(\alpha)$ is not a perfect power, then $\alpha$ generates $K$ over $\Bbb{Q}$.
Proof: The norm $N_{K/\Bbb{Q}}(\alpha)$ is the constant term of the characteristic polynomial $\chi_{K/\Bbb{Q}}(\alpha)\in \Bbb{Z}[X]$, which can be shown to be a power of the minimal polynomial of $\alpha$.


Note 1:You can compute $N_{K/\Bbb{Q}}(\sqrt{2}+\sqrt[3]{5})$ as either
$$
\prod_{i = 0}^2 (\sqrt{2}+\zeta_3^i\sqrt[3]{5})(-\sqrt{2}+\zeta_3^i\sqrt[3]{5})
\quad \text{or} \quad
\det(R_{\sqrt{2}+\sqrt[3]{5}})
$$
where $\zeta_3$ is a primitive third root of unity and $R_{\sqrt{2}+\sqrt[3]{5}}$ is the matrix $(r_{ij})$ with
$$
m_i \cdot (\sqrt{2}+\sqrt[3]{5}) = \sum_j r_{ij} m_j
$$
for every element $m_i,m_j$ of a $\Bbb{Z}$-basis of $\mathcal{O}_K$ (for example $\{1, \sqrt{2}, \sqrt[3]{5}, (\sqrt[3]{5})^2, \sqrt{2}\sqrt[3]{5}, \sqrt{2}(\sqrt[3]{5})^2\}$).

Note 2: The characteristic (hence, as we've seen, minimal) polynomial of $\sqrt{2}+\sqrt[3]{5}$ is (by definition) the characteristic polynomial of the matrix $R_{\sqrt{2}+\sqrt[3]{5}}$, i.e. $f(X)=\det(XI - R_{\sqrt{2}+\sqrt[3]{5}})$.
A: I'm afraid I can't think of any methods other than brute force.  If we set $\alpha = \sqrt{2} + \sqrt[3]{5}$, then all powers of $\alpha$ lie in the span over $\Bbb Q$ of the set $\{1, \sqrt{2}, \sqrt[3]{5}, \sqrt[3]{25}, \sqrt{2}\sqrt[3]{5}, \sqrt{2}\sqrt[3]{25}\}.$  What we essentially need to show is that the set spanned by the powers of $\alpha$ then contains $6$ linearly independent elements of this span.
I start by calculating $\alpha^2 = 2 + \sqrt[3]{25} + 2\sqrt{2}\sqrt[3]{5}$.  You could then compute $\alpha^3, \alpha^4, \alpha^5$ similarly, and check by row reduction that the resulting set is linearly independent.  
If you're doing computations by hand, it might be somewhat easier, to compute $\alpha(\alpha^2-2) = (\sqrt{2} + \sqrt[3]{5})(\sqrt[3]{25} + 2\sqrt{2}\sqrt[3]{5}) = 5 + 4\sqrt[3]{5} + 3\sqrt{2}\sqrt[3]{25}$, and then similarly compute $\alpha(\alpha(\alpha^2-2)-5)$, and so forth.
A: $\newcommand{\QQ}{{\mathbb Q}}$
Maybe its unfair to use computer algebra on that, but with the three equations $x^2 - 2 = 0$, $y^3 - 5 = 0$ and $t-(x+y) = 0$ and Maple's
eliminate function for eliminating $x,y$ from these three equations I get:
$$[ \left\{ x={\frac {{t}^{3}+6\,t-5}{3\,{t}^{2}+2}},y=-{\frac {-5-2\,{t
}^{3}+4\,t}{3\,{t}^{2}+2}} \right\} , \left\{ {t}^{6}+12\,{t}^{2}-6\,{
t}^{4}+17-10\,{t}^{3}-60\,t \right\} ]
$$
With 'factor' one 'proves' that the polynomial $f(t)$ at the right is irreducible and therefore the minimal polynomial of $\sqrt{2} + 5^{1/3}$. This guarantees the well definedness of the expressions at the left. With a gcd calculation one could invert $3 t^2 + 2$ modulo $f(t)$ and express $x,y$ polynomially in $t$.
Note also that $\QQ(\sqrt{2})$ and $\QQ(5^{1/3})$ are linear disjoint over $\QQ$ and therefore $\dim_\QQ \QQ(\sqrt{2},5^{1/3}) = 6$. As our minimal polynomial of $t = \sqrt{2} + 5^{1/3}$ is of degree $6$ we have another proof of the equality of both fields without expressing $x$ and $y$ by $t$.
A: The minimal polynomial of $\sqrt2+\sqrt[3]5$ is $x^6-6 x^4-10 x^3+12 x^2-60 x+17$.
Since $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ has degree $6$, we must have $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$.
