$$
\begin{aligned}
& \frac{x^3}{(2 x-1)(x+2)(x-3)}=\frac{A}{1} \\
& +\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}
\end{aligned}
$$
$$
\begin{aligned}
& A(2 x-1)(x+2)(x-3)+B(x+2)(x-3) \\
& =\frac{+C(x-3)(2 x-1)+D(2 x-1)(x+2)}{(2 x-1)(x+2)(x-3)} \\
&
\end{aligned}
$$
$$
\begin{gathered}
\therefore A(2 x-1)(x+2)(x-3)+B(x+2) \\
(x-3)+C(x-3)(2 x-1)+D(2 x-1) \\
(x+2)=x^3
\end{gathered}
$$

Put $x=3$, we obtain
$$
D(5)(5)=27 \Rightarrow D=27 / 25
$$

Put $x=-2$, we obtain
$$
C(-5)(-5)=-8 \Rightarrow C=-8 / 25
$$

Put $x=\frac{1}{2}$, we obtain
$$
B(5 / 2)(-5 / 2)=1 / 8 \Rightarrow B=-1 / 50
$$

Put $x=0$, we obtain
$$
\begin{array}{r}
A(-1)(2)(-3)+\left(-\frac{1}{50}\right)(2)(-3)+ \\
\left(-\frac{8}{25}\right)(-3)(-1)+\frac{27}{25}(-1)(2)=0 \\
6 A+\frac{3}{25}-\frac{24}{25}-\frac{54}{25}=0
\end{array}
$$

$$
\begin{array}{ll}
\Rightarrow & 6 A=\frac{75}{25}=3 \\
\Rightarrow & A=\frac{3}{6}=\frac{1}{2}
\end{array}
$$

$$
\begin{aligned}