8.27 Let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots$ satisfy $a_{0}=a_{1}=1$ and
$$\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}, n=2,3, \cdots$$

Find the general term formula of the sequence.
[Solution] Since when $n \geqslant 2$,
$$\sqrt{a_{n}}=\sqrt{a_{n-1}}+\frac{2 a_{n-1}}{\sqrt{a_{n-2}}}$$

Let
$$\begin{array}{l}
b_{n}=\sqrt{\frac{a_{n}}{a_{n-1}}} \text {, we get } b_{1}=1, \text { and } \\
b_{n}=1+2 b_{n-1}, n=2,3, \cdots
\end{array}$$

It is easy to obtain
$$b_{n}=2^{n}-1, n=1,2, \cdots$$

Thus, we have
$$\sqrt{a_{n}}=\left(2^{n}-1\right) \sqrt{a_{n-1}},$$

Therefore, the general term formula is
$$a_{n}=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}, n=1,2,3, \cdots$$