6.152. $\frac{z+4}{z-1}+\frac{z-4}{z+1}=\frac{z+8}{z-2}+\frac{z-8}{z+2}+6$.
Solution.

Domain of definition: $\left\{\begin{array}{l}z \neq \pm 1, \\ z \neq \pm 2 .\end{array}\right.$

From the condition we get:

$$
\begin{aligned}
& \frac{(z+4)(z+1)+(z-4)(z-1)}{(z-1)(z+1)}=\frac{(z+8)(z+2)+(z-8)(z-2)}{(z-2)(z+2)}+6 \Leftrightarrow \\
& \Leftrightarrow \frac{2 z^{2}+8}{z^{2}-1}=\frac{2 z^{2}+32}{z^{2}-4}+6 \Leftrightarrow \frac{z^{2}+4}{z^{2}-1}=\frac{z^{2}+16}{z^{2}-4}+3 \Leftrightarrow \\
& \Leftrightarrow \frac{z^{2}+4}{z^{2}-1}-\frac{z^{2}+16}{z^{2}-4}=3 \Leftrightarrow \frac{\left(z^{2}+4\right)\left(z^{2}-4\right)-\left(z^{2}+16\right)\left(z^{2}-1\right)}{\left(z^{2}-1\right)\left(z^{2}-4\right)}=3 \Leftrightarrow \\
& \Leftrightarrow \frac{-15 z^{2}}{z^{4}-5 z^{2}+4}=3 \Leftrightarrow \frac{5 z^{2}}{z^{4}-5 z^{2}+4}=-1 \Leftrightarrow \frac{5 z^{2}}{z^{4}-5 z^{2}+4}+1=0 \Leftrightarrow \\
& \Leftrightarrow \frac{5 z^{2}+z^{4}-5 z^{2}+4}{z^{4}-5 z^{2}+4}=0, \frac{z^{4}+4}{z^{4}-5 z^{2}+4}=0,
\end{aligned}
$$

from which $z^{4}+4=0$ and there are no solutions.

Answer: no roots.