10. (20 points) Solve the equation in the set of real numbers
$$
\sqrt{3}\left(x^{2}-\frac{1}{x^{2}}\right)-4\left(x+\frac{1}{x}\right)=8
$$
10. Let $x=\tan \theta\left(\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right)$. Then the equation can be transformed into
$$
\begin{array}{l}
\sqrt{3}\left(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\right)-4\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)=8 \\
\Rightarrow \frac{-\sqrt{3} \cos 2 \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}-\frac{4}{\sin \theta \cdot \cos \theta}=8 \\
\Rightarrow-\sqrt{3} \cos 2 \theta-2 \sin 2 \theta=2 \sin ^{2} 2 \theta \\
\Rightarrow 2 \sin 2 \theta(\sin 2 \theta+1)+\sqrt{3} \cos 2 \theta=0 \\
\Rightarrow 4 \sin ^{2} 2 \theta(\sin 2 \theta+1)^{2}-3 \cos ^{2} 2 \theta=0 \\
\Rightarrow 4 \sin ^{2} 2 \theta(1+\sin 2 \theta)^{2}-3\left(1-\sin ^{2} 2 \theta\right)=0 .
\end{array}
$$

Let $\sin 2 \theta=y$. Then
$$
y=\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\frac{2 x}{1+x^{2}},
$$

and the equation can be transformed into
$$
\begin{array}{l}
(y+1)(2 y-1)\left(2 y^{2}+3 y+3\right)=0 \\
\Rightarrow y=-1 \text { or } y=\frac{1}{2} .
\end{array}
$$

When $y=\frac{2 x}{1+x^{2}}=-1$, $x=-1$;
When $y=\frac{2 x}{1+x^{2}}=\frac{1}{2}$, $x=2 \pm \sqrt{3}$.
Upon verification, the roots of the original equation are
$$
x_{1}=-1, x_{2}=2+\sqrt{3} .
$$