How would you solve this polynomial? Is there a way to find the roots of equations such as $x^3-9\sqrt[3]{2}+9=0$?
I've just been using Wolfram Alpha to factor it into $(x-\sqrt[3]{4}+\sqrt[3]{2}-1)(x^2+(1-\sqrt[3]{2}+\sqrt[3]{4})x+3\sqrt[3]{4}-3)$. But for harder equations such as $$x^3-63\sqrt[3]{20}+9=0$$, Wolfram Alpha just factors it into $$(x-\text{root of }x^9+27x^6+243x^3-5000211)(x-\text{same thing})(x-\text{same thing})$$
Which is kind of problematic, because I would like the exact values of the factors.
So is there a way to factor such polynomials into multiple factors? Or some way to find its roots! Anything helps.

Note: I would not like nested radicals such as $x=\sqrt[3]{3-\sqrt[4]{4}}$ because I then have to go through the process of finding there simplified surds.

We can find $\sqrt[3]{9(\sqrt[3]2-1)}$ by the following way without WA.
Indeed, let $\sqrt[3]2=x$.
Hence, $$x^3=2$$ or $$9(x^3-1)=9$$ or $$9(x-1)=\frac{9}{1+x+x^2}$$ or $$9(x-1)=\frac{27}{x^3+3x^2+3x+1}$$ or
$$\sqrt[3]{9(\sqrt[3]2-1)}=\frac{3}{\sqrt[3]2+1}$$ or
$$\sqrt[3]{9(\sqrt[3]2-1)}=\sqrt[3]4-\sqrt[3]2+1$$
and we are done!
