5. (5 points) Solve the system of equations:

$$
\left\{\begin{array}{c}
x^{3}+2 y=y^{3}+14 x \\
\frac{y^{2}+1}{x^{2}+1}=3
\end{array}\right.
$$
# Solution:

The following equivalent transformations are valid:

$$
\left\{\begin{array} { c } 
{ x ^ { 3 } + 2 y = y ^ { 3 } + 1 4 x } \\
{ \frac { y ^ { 2 } + 1 } { x ^ { 2 } + 1 } = 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y^{2}=3 x^{2}+2 \\
x^{3}+2 y=y^{3}+14 x
\end{array} \Leftrightarrow\right.\right.
$$

$$
\begin{aligned}
& {\left[\begin{array}{c}
y=\sqrt{3 x^{2}+2} \\
\left\{\begin{array}{c}
y \\
x^{3}+2 \sqrt{3 x^{2}+2}=\left(3 x^{2}+2\right) \sqrt{3 x^{2}+2}+14 x
\end{array}\right. \\
\left\{\begin{array}{c}
3 x^{2}+2 \\
x^{3}-2 \sqrt{3 x^{2}+2}=-\left(3 x^{2}+2\right) \sqrt{3 x^{2}+2}+14 x
\end{array}\right.
\end{array} \Leftrightarrow\right.} \\
& {\left[\begin{array} { c } 
{ y = \sqrt { 3 x ^ { 2 } + 2 } } \\
{ x ^ { 3 } = 3 x ^ { 2 } \sqrt { 3 x ^ { 2 } + 2 } + 1 4 x } \\
{ y = - \sqrt { 3 x ^ { 2 } + 2 } } \\
{ x ^ { 3 } = - 3 x ^ { 2 } \sqrt { 3 x ^ { 2 } + 2 } + 1 4 x }
\end{array} \Leftrightarrow \left[\begin{array}{c}
\left\{\begin{array}{c}
x=0 \\
y= \pm \sqrt{2}
\end{array}\right. \\
\left\{\begin{array}{c}
y=\sqrt{3 x^{2}+2} \\
x^{2}=3 x \sqrt{3 x^{2}+2}+14
\end{array}\right. \\
\left\{\begin{array}{c}
y=-\sqrt{3 x^{2}+2} \\
x^{2}=-3 x \sqrt{3 x^{2}+2}+14
\end{array}\right.
\end{array} \Leftrightarrow\right.\right.} \\
& {\left[\begin{array}{c}
\left\{\begin{array}{c}
x=0 \\
y= \pm \sqrt{2}
\end{array}\right. \\
\left\{\begin{array}{c}
y=\sqrt{3 x^{2}+2} \\
\sqrt{3 x^{2}+2}=\frac{x^{2}-14}{3 x}
\end{array}\right. \\
\left\{\begin{array}{c}
y=-\sqrt{3 x^{2}+2} \\
\sqrt{3 x^{2}+2}=-\frac{x^{2}-14}{3 x}
\end{array}\right.
\end{array}\right.}
\end{aligned}
$$

$$
\left[\begin{array}{c}
\left\{\begin{array}{c}
x=0 \\
y= \pm \sqrt{2}
\end{array}\right. \\
\left\{\begin{array}{c}
y=\sqrt{3 x^{2}+2} \\
\frac{x^{2}-14}{3 x} \geq 0
\end{array}\right. \\
3 x^{2}+2=\frac{\left(x^{2}-14\right)^{2}}{9 x^{2}} \\
\left\{\begin{array}{r}
y=-\sqrt{3 x^{2}+2} \\
\frac{x^{2}-14}{3 x} \leq 0
\end{array}\right. \\
3 x^{2}+2=\frac{\left(x^{2}-14\right)^{2}}{9 x^{2}}
\end{array}\right.
$$

Solve the equation $3 x^{2}+2=\frac{\left(x^{2}-14\right)^{2}}{9 x^{2}}$ for $x \neq 0$. This equation is equivalent to the biquadratic equation $13 x^{4}+23 x^{2}-88=0$, whose solutions are $x=\sqrt{2}$ and $x=-\sqrt{2}$.

Substituting the obtained values into the constraints of the system (1), we get the answer.

Answer: $(0 ; \pm \sqrt{2}),( \pm \sqrt{2} ; \mp 2 \sqrt{2})$.