Simplify $\sqrt[3]{9\sqrt3-11\sqrt2}$ Simplify $$\sqrt[3]{9\sqrt3-11\sqrt2}$$
How can we actually simplify this radical?

Starting with the observation that $\,9^2 \cdot 3 - 11^2 \cdot 2 = 1\,$, let $\,a = \sqrt[3]{9\sqrt3-11\sqrt2}\,$, $\,b = \sqrt[3]{9\sqrt3+11\sqrt2}\,$. Then $\require{cancel}\,a^3+b^3=9\sqrt3-\cancel{11\sqrt2}+9\sqrt3+\cancel{11\sqrt2}=18\sqrt{3}\,$ and $\,ab = \sqrt[3]{9^2 \cdot 3 - 11^2 \cdot 2} = 1\,$.
It follows that $\,(a+b)^3 = a^3+b^3+3ab(a+b)=18 \sqrt{3} + 3(a+b)\,$, so $\,t=a+b\,$ satisfies the equation $\,t^3 - 3 t - 18 \sqrt{3}=0\,$. The substitution $\,t = \sqrt{3}\,t'\,$ gives $\,3 \sqrt{3}\, t'^{\,3} - 3 \sqrt{3}\, t' - 18 \sqrt{3} = 0\,$, or $\,t'^{\,3}- t' - 6 = 0 \iff (t' - 2) (t'^{\,2} + 2 t' + 3) = 0\,$ with the only real root $\,t' = 2 \iff t = 2 \sqrt{3}\,$.
Then $\,a+b=2\sqrt{3}\,$, $\,ab=1\,$, so $\,a,b\,$ are the roots of $\,x^2 - 2 \sqrt{3} x + 1 = 0 \iff x = \sqrt{3} \pm \sqrt{2}\,$ with $\,a\,$ being the smaller root i.e. $\,a = \sqrt{3} - \sqrt{2}\,$, $\,b = \sqrt{3} + \sqrt{2}\,$.

More generally, consider the case of $\,a,b = \sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ with $\,m^2 \cdot p - n^2 \cdot q = r^3\,$. Then, similarly to above, $\,a^3+b^3 = 2m\sqrt{p}\,$ and $\,ab = r\,$, so $\,(a+b)^3 = 2m \sqrt{p} + 3r(a+b)\,$, or $\,t^3 - 3r\,t - 2m \sqrt{p} = 0\,$ where $\,t=a+b\,$, or $\,p\, t'^{\,3} - 3r\, t' - 2m = 0\,$ where $\,t' = \frac{1}{\sqrt{p}}\,t\,$.
If the latter cubic has an eligible rational root $\,t' = 2s\,$, then $\,a+b=t=\sqrt{p}\,t'=2s\,\sqrt{p}\,$ and $\,ab = r\,$, so $\,a,b\,$ are the roots of $\,x^2 - 2s\sqrt{p}\, x + r = 0\,$ $\,\iff a,b = s\,\sqrt{p} \pm \sqrt{s^2 p - r}\,$.
Therefore a sufficient condition for $\,a,b = \sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ to denest is for $\,m^2 \cdot p - n^2 \cdot q\,$ to be the cube of a rational $\,r\,$, and for the cubic $\,p\, t'^{\,3} - 3r\, t' - 2m\,$ to have an appropriate rational root, and in that case $\,a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p-4r}\right)\,$.
