6.059. $\sqrt[5]{\frac{16 z}{z-1}}+\sqrt[5]{\frac{z-1}{16 z}}=2.5$.
Solution.

Domain of definition: $\left\{\begin{array}{l}z \neq 1, \\ z \neq 0 .\end{array}\right.$

Let $\sqrt[5]{\frac{16 z}{z-1}}=y, y \neq 0$. The equation in terms of $y$ becomes $y+\frac{1}{y}=2.5 \Leftrightarrow y^{2}-2.5 y+1=0$, from which $y_{1}=\frac{1}{2}, y_{2}=2$. Then either

$\sqrt[5]{\frac{16 z}{z-1}}=\frac{1}{2}, \frac{16 z}{z-1}=\frac{1}{32}, z_{1}=-\frac{1}{511}$, or $\sqrt[5]{\frac{16 z}{z-1}}=2, \frac{16 z}{z-1}=32, z_{2}=2$.

Answer: $z_{1}=-\frac{1}{511}, z_{2}=2$.