2. In $\triangle A B C$, $A B=A C=a, B C=b$, $\angle A=36^{\circ}$. Let $m=\frac{a+b}{a-b}, n=\frac{(a+b)^{2}}{a b}, p=\frac{a^{3}}{b^{3}}$. Then the size relationship of $m, n, p$ is $(\quad$.
(A) $m>n>p$
(B) $n>p>m$
(C) $p>m>n$
(D) $m=n=p$
2.D.

From the problem, we know $\angle A B C=\angle A C B=72^{\circ}$. As shown in Figure 4, construct $B D$ to bisect $\angle A B C$. Then
$$
\begin{array}{l}
\angle A B D=\angle D B C=36^{\circ}, \\
\angle B D C=72^{\circ} .
\end{array}
$$

Therefore, $\triangle B C D \backsim \triangle A B C$
$$
\begin{aligned}
& \frac{C D}{B C}=\frac{B C}{A B} \\
\Rightarrow & \frac{a-b}{b}=\frac{b}{a} \\
\Rightarrow & b^{2}=a(a-b) \Rightarrow b(a+b)=a^{2} \\
\Rightarrow & \frac{a+b}{a-b}=\frac{\frac{a^{2}}{b}}{\frac{b^{2}}{a}}=\frac{a^{3}}{b^{3}} \Rightarrow m=p . \\
& \frac{(a+b)^{2}}{a b}=\frac{\frac{a^{4}}{b^{2}}}{a b}=\frac{a^{3}}{b^{3}}, \text { i.e., } n=p .
\end{aligned}
$$

Thus, $m=n=p$.