Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb Q$ Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$  as a root).
I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both sides. But I keep on going into a loop by continuing to  square it over and over again. Then I tried $a^3=(\sqrt[3]{2} +\sqrt{5})^3$. Just can't seem to get rid of the radicals. 
 A: $x = \sqrt[3]{2} + \sqrt{5}$
$x - \sqrt{5} = \sqrt[3]{2}$
$x^3 - 3\sqrt{5}x^2 + 15x - 5\sqrt{5} = 2$
$x^3 - 15x - 2 = \sqrt{5}(3x^2 + 5)$
$(x^3 - 15x - 2)^2 = 5(3x^2 + 5)^2$ 
$x^6 -75x^4 - 4x^3 - 400x^2 + 60x - 121 = 0$
Now, ignoring accuracy and assuming arithmetical errors are both inevitable and irrelevant, there is another correctly figured out 6th degree polynomial  with $ \sqrt[3]{2} + \sqrt{5}$ as a solution.
Unless, I didn't make an arithmetical error calculating this.  I might not have.  There's a first time for everything.
====
So obviously I made an arithmetical error.  I knew I would.  But it doesn't matter as the solving for the polynomial doesn't affect that there will be such a polynomial.  
$x = \sqrt[3]{2} + \sqrt{5} \iff$
$x + k_1\sqrt{5} = \sqrt[3]{2} \iff$
$x^3 + k_2\sqrt{5}x^2 + k_3*5x + k_4*5*\sqrt{5} = 2 \iff$
$x^3 + k_5x + k_6 = \sqrt{5}(k_7x^2 + k_8) \iff $
$(x^3 + k_5x + k_6)^2= 5(k_7x^2 + k_8)^2 \iff $
$x^6 + k_9x^5 + k_{10}x^4 + k_{11}x^3 + k_{12}x^2 + k_{13}x + k_{14} = 0$
Where $k_i$ are integers and easily calculatable by someone who can count to 20 with his shoes on.
A: $$r-\sqrt5=\sqrt[3]2\implies r^3-3\sqrt5\,r^2+15\,r-5\sqrt5=2\implies$$
$$r^3+15r-2=(3r^2+5)\sqrt5\implies r^6+30r^4+225r^2-4r^3-60r+4=45r^4+150r^2+125$$
Now group up elements and put $\;x=r\;$ and there you have a rational polynomial for which $\;\sqrt[3]2+\sqrt5\;$ is a root.
A: As a vector space over $\mathbb Q$, $\mathbb Q(\sqrt[3]{2},\sqrt 5)$ has a basis $\{1,\sqrt 5,\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{2}\cdot\sqrt 5,\sqrt[3]{4}\cdot\sqrt 5\}$. Now consider the linear map $\phi:\mathbb Q(\sqrt[3]{2},\sqrt 5)\to\mathbb Q(\sqrt[3]{2},\sqrt 5)$ given by $\phi(y)=(\sqrt[3]{2}+\sqrt 5)y$. With respect to this basis, its matrix is given by
$$
\begin{bmatrix}
\phi(1)&\phi(\sqrt 5)&\phi(\sqrt[3]{2})&\phi(\sqrt[3]{4})&\phi(\sqrt[3]{2}\cdot\sqrt 5)&\phi(\sqrt[3]{4}\cdot\sqrt 5)
\end{bmatrix}
$$
which comes out to
$$
A=\begin{bmatrix}
0&5&0&2&0&0\\
1&0&0&0&0&2\\
1&0&0&0&5&0\\
0&0&1&0&0&5\\
0&1&1&0&0&0\\
0&0&0&1&1&0
\end{bmatrix}.
$$
The characteristic polynomial $f(x)=\det(xI-A)$ has $\sqrt[3]{2}+\sqrt 5$ as a root, since $\sqrt[3]{2}+\sqrt 5$ is an eigenvalue of $\phi$. This turns out to be
$$
\det\begin{bmatrix}
x&-5&0&-2&0&0\\
-1&x&0&0&0&-2\\
-1&0&x&0&-5&0\\
0&0&-1&x&0&-5\\
0&-1&-1&0&x&0\\
0&0&0&-1&-1&x
\end{bmatrix}=x^6-15x^4-4x^3+75x^2-60x-121.
$$
A: Hint. Let $a,b$ be algebraic over field $F$ and $m_a(x), m_b(x)\in F[x]$ be minimal polynomials of $a$ and $b$, respectively. Then $a+b, ab$ are algebraic over $F$ and
$$
\operatorname{deg}m_{a+b}(x)\leq \operatorname{deg}m_{a}(x)\operatorname{deg}m_{b}(x); \ \operatorname{deg}m_{a\cdot b}(x)\leq \operatorname{deg}m_{a}(x)\operatorname{deg}m_{b}(x)
$$
