8. The integer part and decimal part of $(\sqrt{10}+3)^{2 n+1}(n \in \mathbf{N})$ are $I$ and $F$, respectively, then the value of $P(I$ $+F)$ is $\qquad$
8. 1

Consider $(\sqrt{10}-3)^{2 n+1}$. Since $(\sqrt{10}-3)^{2 n+1} \cdot(\sqrt{10}+3)^{2 n+1}=1$. Therefore, $(\sqrt{10}-3)^{2 n+1}=\frac{1}{(\sqrt{10}+3)^{2 n+1}}=\frac{1}{1+F}$. By the binomial theorem, $(\sqrt{10}+3)^{2 n+1} - (\sqrt{10}-3)^{2 n+1}$ is an integer. That is, $I+F-\frac{1}{I+F}$ is an integer. Thus, $F-\frac{1}{I+F}$ is an integer. Since $0<F<1, 0<\frac{1}{I+F}<1$, it follows that $F-\frac{1}{I+F}=0$, which means $F(I+F)=1$.