Simplify $\sqrt[3]{9\sqrt3-11\sqrt2}$ Simplify $$\sqrt[3]{9\sqrt3-11\sqrt2}$$
How can we actually simplify this radical?
 A: 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,
$$x = \sqrt 3 - \sqrt 2$$
A: It would seem that the answer is
$$
\sqrt{3}-\sqrt{2}.
$$
See this C# program.
A: 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 $\,a,b = \frac{1}{2}\left(t'\,\sqrt{p} \pm \sqrt{t'^{\,2} p-4r}\right)\,$.
A: Apply the denesting formula
$$\sqrt[3]{a\sqrt d+b \sqrt c}=\frac12\sqrt[3]{\frac{3bt-a}d}\left(\sqrt d+\frac1t \sqrt c\right)$$
with $t$ satisfying $t^3-\frac{3a}bt^2+\frac{3c}d t-\frac{ac}{bd}=0$. Then
$$\sqrt[3]{9\sqrt3-11\sqrt2} = \frac12\sqrt[3]{-11t-3}\left(\sqrt 3+\frac1t \sqrt 2\right) $$
where $t^3+\frac{27}{11}t^2+2t+\frac{6}{11}=0$, yielding $t=-1$. Plug into above expression to obtain
$$\sqrt[3]{9\sqrt3-11\sqrt2}
=\sqrt{3}-\sqrt2$$
