6. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=1, a_{1}=1, a_{n}=a_{n-1}+\frac{a_{n-1}^{2}}{a_{n-2}}(n \geqslant 2)$; the sequence $\left\{b_{n}\right\}$ satisfies $b_{0}=1, b_{1}=3, b_{n}=b_{n-1}+\frac{b_{n-1}^{2}}{b_{n-2}}(n \geqslant 2)$. Then $\frac{b_{32}}{a_{32}}=$ $\qquad$
6. 561.

Let $x_{n}=x_{n-1}+\frac{x_{n-1}^{2}}{x_{n-2}}, y_{n}=\frac{x_{n}}{x_{n-1}}$, then $y_{n}=1+y_{n-1}$. Therefore,
$$
\frac{x_{n}}{x_{0}}=\frac{x_{n}}{x_{n-1}} \cdot \frac{x_{n-1}}{x_{n-2}} \cdots \cdots \frac{x_{1}}{x_{0}}=y_{1} y_{2} \cdots y_{n}=y_{1}\left(y_{1}+1\right) \cdots\left(y_{1}+n-1\right)
$$

In the first sequence, $y_{1}=\frac{a_{1}}{a_{0}}=1$, hence $a_{n}=n!$. Similarly, in the second sequence, $y_{n}=\frac{b_{1}}{b_{0}}=3$, hence $b_{n}=\frac{(n+2)!}{2}$.