1. Solve the equations:
$$
\text { (1) } \begin{aligned}
& (a+b)(a x+b)(a-b x) \\
= & \left(a^{2} x-b^{2}\right)(a+b x) ;
\end{aligned}
$$
(2) $x^{\frac{1}{3}}+(2 x-3)^{\frac{1}{3}}=\{12(x-1)\}^{\frac{1}{3}}$.
(1) It is obvious that $x=1$ is a solution to the equation. Simplify this equation to
$$
\begin{array}{l}
\quad(a+b)\left\{a b+\left(a^{2}-b^{2}\right) x-a b x^{2}\right\} \\
=a^{3} x-a b^{2}+a^{2} b x^{2}-b^{3} x . \\
\quad \text { From the equation }\left[a^{2} b+a b(a+b)\right) x^{2} \\
+\left[a^{3}-i^{2}-(a+i)\left(a^{2}-b^{2}\right)\right] x \\
-\left[a b^{2}+a b(a+b)\right]=0
\end{array}
$$

we can obtain the product of the two roots as
$$
\begin{aligned}
x_{1} \cdot x_{2} & =-\frac{a b^{2}+a b(a+b)}{a^{2} b+a b(a+b)} \\
& =-\frac{a+2 b}{2 a+b} .
\end{aligned}
$$

Thus, the second root can be easily found, which is
$$
-\frac{a+2 b}{2 a+b} \text {. }
$$
(2) Let $c=a+b$. Then,
$$
\begin{aligned}
c^{3} & =(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b) \\
& =a^{3}+b^{3}+3 a b c .
\end{aligned}
$$

Rearranging gives
$$
3 a b c=c^{3}-a^{3}-b^{3} .
$$

Therefore, the equivalent equation can be written as
$$
\begin{aligned}
& 3 \sqrt{12 x}(2 x-3)(x-1) \\
= & 12(x-1)-x-(2 x-3)=9(x-1) .
\end{aligned}
$$

Cubing both sides yields
$$
12 x(2 x-3)(x-1)=27(x-1)^{3} .
$$

Thus,
$$
x-1=0 \text { or } 4 x(2 x-3)=9(x-1)^{2} \text {. }
$$

Therefore, $x=1, x=3$ are the roots of the equation.