minimum polynomial of $\sqrt[3]{3}+\sqrt[3]{5}$ over $\mathbb {Q}$

I have tried putting $x=\sqrt[3]{3}+\sqrt[3]{5}$ but i got stuck because after elevating to the third power I obtain again cube roots and I am stuck . How can I solve?

\begin{align} x &= \sqrt[3]{3} + \sqrt[3]{5} \\ x^3 &= 3 + 3\sqrt[3]{3\cdot 3 \cdot 5} + 3\sqrt[3]{3 \cdot 5 \cdot 5} + 5 \\ x^3 - 8 &= 3\sqrt[3]{3 \cdot 5}(\sqrt[3]{3} + \sqrt[3]{5}) \\ x^3 - 8 &= 3\sqrt[3]{15}x\\ (x^3-8)^3 &= 405 x^3 \\ x^9 - 24x^6 - 213 x^3 - 512 &= 0. \end{align} Therefore the minimum polynomial of $\sqrt[3]{3} + \sqrt[3]{5}$ is a factor of $x^9 - 24x^6 - 213 x^3 - 512$.

If shouldn't be difficult to prove that the defining field is of degree 9, and therefore this is the minimal polynomial.