Example 3 Let $a$ and $b$ be positive real numbers, and
$$
\begin{array}{l}
\frac{1}{a}-\frac{1}{b}-\frac{1}{a+b}=0 . \\
\text { Then }\left(\frac{b}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3}=
\end{array}
$$
Given that $\frac{b}{a}-\frac{a}{b}=1$.
$$
\text { Let } \frac{b}{a}=\frac{1}{2}+t, -\frac{a}{b}=\frac{1}{2}-t \text {. }
$$

Multiplying the two equations gives $\frac{1}{4}-t^{2}=-1$.
Solving for $t$ yields $t= \pm \frac{\sqrt{5}}{2}$.
Since $a$ and $b$ are both positive real numbers, then
$$
\frac{b}{a}=\frac{1}{2}+t>0 \text {. }
$$

Therefore, $t=\frac{\sqrt{5}}{2}$.
Thus, $\frac{b}{a}+\frac{a}{b}=2 t=\sqrt{5}$.
$$
\begin{array}{l}
\text { Hence }\left(\frac{b}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3} \\
=\left(\frac{b}{a}+\frac{a}{b}\right)\left[\left(\frac{b}{a}+\frac{a}{b}\right)^{2}-3 \times \frac{b}{a} \times \frac{a}{b}\right] \\
=\sqrt{5}\left[(\sqrt{5})^{2}-3 \times 1\right]=2 \sqrt{5} .
\end{array}
$$