8. Given $x-y=1(x>1)$, then the minimum value of $k=\frac{1}{y^{3}}+x+1$ is

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8. Given $y=x-1$, substituting we get
$$
\begin{aligned}
k & =\frac{1}{(x-1)^{3}}+x+1=\frac{1}{(x-1)^{3}}+\frac{x-1}{3}+\frac{x-1}{3}+\frac{x-1}{3}+2 \\
& \geqslant 4 \sqrt[1]{\frac{1}{(x-1)^{3}} \cdot\left(\frac{x-1}{3}\right)^{3}}+2=\frac{4}{3} \sqrt[3]{3}+2 .
\end{aligned}
$$

When $\frac{1}{(x-1)^{3}}=\frac{x-1}{3}$, i.e., $x=1+\sqrt[1]{3}$, $k$ takes the minimum value $\frac{4}{3} \sqrt[1]{3}+2$.