Showing that $\sqrt{2} + \sqrt[3]{5}$ is algebraic of degree $6$ over $\mathbb{Q}$ 
Show that $\sqrt{2} + \sqrt[3]{5}$ is algebraic of degree $6$ over
  $\mathbb{Q}$

What is the degree of a root? Is it the smallest polynomial that gives this thing as root?
What I tried:
$x = \sqrt{2} + \sqrt[3]{5} \implies x^2 = 2 + 2\sqrt{2}\sqrt[3]{5} + \sqrt[3]{5^2}$
I don't see it going anywhere.
Maybe if I try to elevate to the power of something that is common to both $2$ and $3$, as $6$, I'll get something. But elevating the root itself won't help: http://www.wolframalpha.com/input/?i=(%5Csqrt%7B2%7D+%2B+%5Csqrt%5B3%5D%7B5%7D)%5E6
Do I really need to find the polynomial or just show that it is algebraic using another technique?
I couldn't find anything helpful here The degree of $\sqrt{2} + \sqrt[3]{5}$ over $\mathbb Q$
 A: It is easily seen that a sum of two algebraic numbers $\alpha$ and $\beta$ is algebraic since one has:
$$\mathbb{Q}(\alpha+\beta)\subseteq\mathbb{Q}(\alpha,\beta).$$
Hence, $\mathbb{Q}(\alpha+\beta)$ is finite-dimensional since $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)(\beta)$ is, indeed recall that:
$$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}].$$

If $p$ is an annihilator polynomial of $\alpha$ and $q$ one of $\beta$, then $\textrm{res}_Y(p(Y),q(X-Y))$ is an annihilator polynomial of $\alpha+\beta$.

In your case, use $p=X^2-2$ and $q=X^3-5$ to find that $X^6-6X^4-10X^3+12X^2-60X+17$ is an annihilator polynomial of $\sqrt{2}+\sqrt[3]{5}$. You are left to show that it is indeed irreducible over $\mathbb{Q}$.
A: Hint: Use the formula for the degree:
$[\mathbb{Q}(\sqrt{2}+\sqrt[3]{5}):\mathbb{Q}]= [\mathbb{Q}(\sqrt{2}+\sqrt[3]{5}):\mathbb{Q}(\sqrt{2})]\cdot [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$
A: $$\sqrt[3]5+\sqrt2=q\implies$$
$$q-\sqrt2=\sqrt[3]5\implies$$
$$q^3-3\sqrt2q^2+6q-2\sqrt2=5\implies$$
$$q^3+6q+5=(3q^2+2)\sqrt2.$$
Square the last identity.
