6.235. $\left\{\begin{array}{l}\sqrt[3]{\frac{y+1}{x}}-2 \sqrt[3]{\frac{x}{y+1}}=1, \\ \sqrt{x+y+1}+\sqrt{x-y+10}=5 .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq-1, \\ x+y+1 \geq 0, \\ x-y+10 \geq 0 .\end{array}\right.$

Rewrite the first equation of the system as

$\sqrt[3]{\frac{y+1}{x}}-\frac{2}{\sqrt[3]{\frac{y+1}{x}}}-1=0 \Rightarrow\left(\sqrt[3]{\frac{y+1}{x}}\right)^{2}-\sqrt[3]{\frac{y+1}{x}}-2=0$. Solving this equation as a quadratic equation in terms of $\sqrt[3]{\frac{y+1}{x}}$, we have $\sqrt[3]{\frac{y+1}{x}}=-1$, from which $\frac{y+1}{x}=-1, y=-x-1$ or $\sqrt[3]{\frac{y+1}{x}}=2$, from which $\frac{y+1}{x}=8, y=8 x-1$. From the second equation of the system $\sqrt{x-x-1+1}+\sqrt{x+x+1+10}=5 \Leftrightarrow \sqrt{2 x+11}=5$, $\sqrt{x+8 x-1+1}+\sqrt{x-8 x+1+10}=5 \Leftrightarrow \sqrt{9 x}+\sqrt{-7 x+11}=5 . \quad x_{1}=7, \quad x_{2}=\frac{49}{64}$, $x_{3}=1$. Then $y_{1}=-8, y_{2}=\frac{41}{8}, y_{3}=7$.

Answer: $(7 ;-8),\left(\frac{49}{64} ; \frac{41}{8}\right),(1 ; 7)$.