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

Let $x = \sqrt[3]{9\sqrt 3 -11 \sqrt 2}$.  Cubing both sides gives $x^3 = 9\sqrt 3 -11 \sqrt 2$.
Per the suggestions in the comments, assume $x = a\sqrt 3 + b \sqrt 2$.  Then:
$$x^3 = (a\sqrt 3 + b \sqrt 2)^3$$
$$x^3 = (a\sqrt 3)^3 + 3(a\sqrt 3)^2(b \sqrt 2) + 3(a\sqrt 3)(b \sqrt 2)^2 + (b \sqrt 2)^3$$
$$x^3 = a^3(\sqrt 3)^3 + 3a^2b(\sqrt 3)^2(\sqrt 2) + 3ab^2(\sqrt 3)(\sqrt 2)^2 + b^3(\sqrt 2)^3$$
$$x^3 = 3a^3\sqrt 3 + 9a^2b \sqrt 2 + 6ab^2 \sqrt 3 + 2 b^3 \sqrt 2$$
$$x^3 = (3a^3 + 6ab^2)\sqrt 3 + (9a^2b + 2b^3)\sqrt 2$$
If we can find $a$ and $b$ such that $3a^3 + 6ab^2 = 9$ and $9a^2b + 2b^3 = -11$, then it will solve the original equation.
From $3a^3 + 6ab^2 = 9$, we get $b = \pm \sqrt{\frac{3 - a^3}{2a}}$.  Plugging this into the other equation gives:
$$9a^2(\pm \sqrt{\frac{3 - a^3}{2a}}) + 2(\pm \sqrt{\frac{3 - a^3}{2a}})^3 = -11$$
which, with a bunch of algebra that I won't show here, simplifies to:
$$64a^6 - 73a^3 + 9 = 0$$
Let $u=a^3$.  Then $64u^2 - 73u + 9 = 0$.  Applying the quadratic formula gives:
$$u = \frac{73 \pm \sqrt{3025}}{128} = \frac{73 \pm 55}{128}$$
Which has the solutions $u=1$ or $u=\frac{9}{64}$.
If $u=1$, then $a=1$, so $b=\pm 1$.
If $u=\frac{9}{64}$, then $a=\frac{\sqrt[3]9}{4}$, and $b=\pm\sqrt[6]{\frac{680943}{2^{15}}}$.
That gives 4 possible combinations for $a$ and $b$, and it just so happens that $a=1$ and $b=-1$ satisfies the original equation.  Therefore,
$$x = \sqrt 3 - \sqrt 2$$
