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. 

$x = \sqrt[3]{2} + \sqrt{5}$
$x - \sqrt{5} = \sqrt[3]{2}$
$x^3 - 3\sqrt{5}x^2 + 15x - 5\sqrt{5} = 2$
$x^3 - 15x - 2 = \sqrt{5}(3x^2 + 5)$
$(x^3 - 15x - 2)^2 = 5(3x^2 + 5)^2$ 
$x^6 -75x^4 - 4x^3 - 400x^2 + 60x - 121 = 0$
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.
