Example 2 Given that $a$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Then the value of $\frac{a^{3}-1}{a^{5}+a^{4}-a^{3}-a^{2}}$ is $\qquad$ .
(1995, National Junior High School Mathematics League)
Solution: The method of solving for the root $a$ by setting up an equation and then substituting to find the value is too cumbersome. It is better to use an overall processing method for this problem. From the given, we have
$$
a^{2}+a=\frac{1}{4} \text { or } a^{2}=\frac{1}{4}-a \text {. }
$$

Using equation (1), we simplify the numerator and denominator of the expression to be evaluated respectively:
$$
\begin{array}{l}
a^{3}-1=(a-1)\left(a^{2}+a+1\right) \\
\quad=(a-1)\left(\frac{1}{4}+1\right)=\frac{5}{4}(a-1) ; \\
a^{5}+a^{4}-a^{3}-a^{2} \\
=a^{3}\left(a^{2}+a\right)-a\left(a^{2}+a\right) \\
=\frac{1}{4} a^{3}-\frac{1}{4} a \\
=\frac{1}{4}\left[\frac{5}{4}(a-1)+1\right)-\frac{1}{4} a \\
=\frac{1}{16}(a-1) .
\end{array}
$$

Therefore, the original expression $=20$.