8. Given that $a, b, x, y$ are all positive numbers, $n$ is a positive integer, and $x+y=1$, find the minimum value of $\frac{a}{x^{n}}+\frac{b}{y^{n}}$.
8. For any real number $\lambda$, we have
$$
\begin{aligned}
& \frac{a}{x^{n}}+\frac{b}{y^{n}}=\frac{a}{x^{n}}+\frac{b}{y^{n}}+n \lambda(x+y)-n \lambda=\frac{a}{x^{n}}+\underbrace{\lambda x+\cdots+\lambda x}_{n \uparrow \lambda x}-n \lambda \\
\geqslant & (n+1) \sqrt[n+1]{\frac{a}{x^{n}} \cdot(\lambda x)^{n}}+(n+1) \sqrt[n+1]{\frac{b}{y^{n}} \cdot(\lambda y)^{n}}-n \lambda \\
= & (n+1)\left[\left(a \lambda^{n}\right)^{\frac{1}{n+1}}+\left(b \lambda^{n}\right) \frac{1}{n+1}\right]-n \lambda .
\end{aligned}
$$

Equality holds if and only if $x=\sqrt[n+1]{\frac{a}{\lambda}}, y=\sqrt[n+1]{\frac{b}{\lambda}}$. In this case, from $x+y=1$, we get $\lambda=(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}, x=$ $\frac{\sqrt[n+1]{a}}{\sqrt[n+1]{a}+\sqrt[n+1]{b}}, y=\frac{\sqrt[n+1]{b}}{\sqrt[n+1]{a}+\sqrt[n+1]{b}}$. Therefore, the minimum value of $\frac{a}{x^{n}}+\frac{b}{y^{n}}$ is $(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}$.