[Example 5.2.1] Solve the equation;
$$
8 x^{3}+8 \sqrt{3} x^{2}+6 x+\sqrt{3}-1=0 .
$$
The unknown in this equation is $x$, and it is a cubic equation in $x$. Solving a cubic equation can be challenging. Can we change our perspective? Temporarily treat the unknown $x$ as a known number, and treat the known number $\sqrt{3}$ as the unknown.
Let $\sqrt{3}=t$, then $t^{2}=3, 2 t^{2}=6$.
At this point, the given equation can be transformed into an equation in terms of $t$:
$$
2 x t^{2}+\left(8 x^{2}+1\right) t+8 x^{3}-1=0.
$$

From the given equation, we know that $x \neq 0$, so the above equation is a quadratic equation in terms of $t$. Calculate its discriminant:
$$
\begin{aligned}
\Delta & =\left(8 x^{2}+1\right)^{2}-8 x\left(8 x^{3}-1\right) \\
& =64 x^{4}+16 x^{2}+1-64 x^{3}+8 x \\
& =16 x^{2}+8 x+1 \\
& =(4 x+1)^{2}.
\end{aligned}
$$

Using the quadratic formula, we get:
$$
\begin{array}{c}
t=\frac{-8 x^{2}-1 \pm(4 x+1)}{4 x} \\
t_{1}=-2 x+1, t_{2}=-\frac{4 x^{2}+2 x+1}{2 x}.
\end{array}
$$

Now, substitute $t=\sqrt{3}$ back:
$$
\begin{array}{c}
\sqrt{3}=-2 x+1, \text{ then } x_{1}=\frac{1-\sqrt{3}}{2}. \\
\sqrt{3}=-\frac{4 x^{2}+2 x+1}{2 x}, \text{ then } x_{2,3}=\frac{-(1+\sqrt{3}) \pm \sqrt[4]{12}}{4}.
\end{array}
$$

Thus, we obtain the three solutions of the equation:
$$
x_{1}=\frac{1-\sqrt{3}}{2}, x_{2}=\frac{-(1+\sqrt{3})+\sqrt[4]{12}}{2}, x_{3}=-\frac{1+\sqrt{3}+\sqrt[4]{12}}{2}.
$$

From the process of solving this equation, we can see that $x$ is the unknown in a cubic equation, but we treated $x$ as a known number, and $\sqrt{3}$, which is a known number, was treated as the unknown. This "reversal" approach transforms a cubic equation into a quadratic equation, thereby converting a difficult equation into a quadratic equation that can be solved using a formula.