8.5. Find the non-zero natural numbers $x, y, z$, knowing that $\frac{x^{3}}{x+26}=\frac{y}{y+25}=\frac{z^{2}}{z+12}$.
Solution: Since $x, y, z$ are non-zero natural numbers, we have:

$y<y+25 \Rightarrow \frac{y}{y+25}<1 \Rightarrow z^{2}<z+12 \Rightarrow z \in\{1,2,3\}$.

$z=1 \Rightarrow \frac{y}{y+25}=\frac{1}{13} \Leftrightarrow 13 y=y+25 \Leftrightarrow y=\frac{25}{12}$ and $y \notin \mathbb{N}$.

$z=2 \Rightarrow \frac{y}{y+25}=\frac{2}{7} \Leftrightarrow 7 y=2 y+50 \Leftrightarrow y=10$ and $x=2$. The argument for the value of $x$:

$\frac{x^{3}}{x+26}=\frac{2}{7} \Leftrightarrow 7 x^{3}=2 x+52 \Leftrightarrow 7\left(x^{3}-8\right)-2(x-2)=0 \Leftrightarrow(x-2)\left(7 x^{2}+14 x+\right.$ 26) $=0 \Leftrightarrow x=2$.

$z=3 \Rightarrow \frac{y}{y+25}=\frac{3}{5} \Leftrightarrow 5 y=3 y+75 \Leftrightarrow y=\frac{75}{2}$ and $y \notin \mathbb{N}$.

Answer: $x=2, y=10, z=2$.