Find the minimal polynomial of $\sqrt[3]{3} + \sqrt{5}$ over $\mathbb{Q}$. Let $x = \sqrt[3]{3} + \sqrt{5}$
Notice that $(x+y)^2 = x^2 + 2xy + y^2$
Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$
Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$
Or notice that $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$
Then, $x^3 = (\sqrt[3]{3} + \sqrt{5})^3 = 3 + 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Then, $x^3 - 3 = 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Which I think the degree of the minial polynomial of $x$ over $\mathbb{Q}$ should have a degree of $6$.
Which means the minimal polynomial can be written as $x^6 + a_5 x^5 + \cdots + a_0$ for some $a_5, ..., a_0 \in \mathbb{Q}$.
But I just don't how get it from both the ways I provide above.
Or that $\sqrt[3]{3} + \sqrt{5}$ is not algebraic. So, I can not write such minimal polynomial out algebraic ?

If $x=\sqrt[3]3+\sqrt5$, then $\left(x-\sqrt5\right)^3=3$. In other words $x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0$. But\begin{align}x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0&\iff x^3+15x-3=(3x^2-5)\sqrt5\\&\implies(x^3+15x-3)^2=5(3x^2-5)^2\\&\iff x^6-15 x^4-6 x^3+375 x^2-90 x-116=0.\end{align}
