Solving equation involving roots and powers . I'm trying to solve this equation :
$\sqrt{3}\sqrt{237x^2 + \frac{224}{x^2}}x^7 + \frac{35293}{222}x^8 + \frac{2}{999}\sqrt{3}{(\sqrt{237x^2 + \frac{224}{x^2}})}^3x^5 - \frac{44968}{111}x^4 + \frac{12544}{333}=0$.
It looks pretty complicated to me, but the computer gives me exact solutions. For example one real solution is $\frac{2}{\sqrt{3}}$, and another : $-\frac{2}{3^{\frac{3}{4}}\sqrt[4]{7}}$ and so on.
Does anyone know how to obtain the exact solutions to the above equation? Thanks in advance!
EDIT:
This came from fooling around with functions that describe surfaces spanned by the roots of polynomials $x^4+c_2x^2+c_3x+c_4$ in 3 dimensions. So here $r_1+r_2+r_3+r_4=0$. Newton's identity $r_1^4+r_2^4+r_3^4+r_4^4 = 4$ (the 4 here is randomly chosen) can thus be represented as a 2D surface in 3D. With some other transformations this gives :
$-\frac{1}{3}\sqrt{6}\sqrt{3}x_1x_2^2x_3 + \frac{1}{9}\sqrt{6}\sqrt{3}x_1x_3^3 + \frac{7}{12}x_1^4 + \frac{1}{2}x_1^2x_2^2 + \frac{1}{2}x_2^4 + \frac{1}{2}x_1^2x_3^2 + x_2^2x_3^2 + \frac{1}{2}x_3^4 = 4$ .
This surface looks somewhat like an octahedron. To find the extrema on the surface I used Lagrange multipliers leading to the equation in question. So I guess it's not very surprising that this has an exact solution. But I was surprised that the computer could find them so easily..
 A: 
how to obtain the exact solutions to the above equation?

Not something you'd want to do by hand, but the equation can be reduced to a quartic in $\,x^4\,$, which just "happens" to factor nicely.
Let $\,y = \sqrt{237x^2 + \dfrac{224}{x^2}}\,$ then, after eliminating the denominators, the original equation can be written as the following system (with the restriction $\,y \ge 0\,$ for the real solutions):
$$
\begin{cases}
\begin{align}
p(x,y) &= 317637\,x^8 + 1998\sqrt{3}\,x^7 y + 4\sqrt{3}\,x^5 y^3 - 809424\,x^4 + 75264 &= 0
\\ q(x,y) &= 237\,x^4 - x^2y^2 + 224 &= 0
\end{align}
\end{cases}
$$
Eliminating $\,y\,$ between the two equations using resultants gives the following, courtesy WA:
$$
\begin{align}
0 = \text{res}_y(p,q) &= -7203 x^6 (13150431 x^{16} - 72718560 x^{12} + 97023744 x^8 \\ &\quad\quad\quad\quad\quad - 16990208 x^4 + 786432)
\\ &= -7203 x^6 (3 x^2 - 4) (3 x^2 + 4) (9 x^4 - 32) (189 x^4 - 16) (859 x^4 - 96)
\end{align}
$$

[ EDIT ] $\;$ The same result can be derived directly by eliminating the radicals from the original equation (just move the two radicals to one side, then square the equation and collect the terms), but the calculations are laborious, which is what I meant by "wouldn't want to do it by hand".
