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. 

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
$$
\det\begin{bmatrix}
x&-5&0&-2&0&0\\
-1&x&0&0&0&-2\\
-1&0&x&0&-5&0\\
0&0&-1&x&0&-5\\
0&-1&-1&0&x&0\\
0&0&0&-1&-1&x
\end{bmatrix}=x^6-15x^4-4x^3+75x^2-60x-121.
$$
