3. Solve the system in the set of real numbers:

$$
x^{2}+x^{2} y^{2}+x^{2} y^{4}=525 \text { and } x+x y+x y^{2}=35 \text {. }
$$
Solution. One

\[
\begin{aligned}
& 525=x^{2}+x^{2} y^{2}+x^{2} y^{4}=\left(x+x y^{2}\right)^{2}-x^{2} y^{2}=\left(x+x y^{2}-x y\right)\left(x+x y^{2}+x y\right) \quad \\
& x+x y+x y^{2}=35 \text { follows that } x-x y+x y^{2}=15 . \text { From } x+x y+x y^{2}=35 \quad \text { and }
\end{aligned}
\]

$x-x y+x y^{2}=15$ it follows that $2 x y=20$, hence $x y=10$. Clearly $x \neq 0$, so if we substitute $y=\frac{10}{x}$ into $x+x y+x y^{2}=35$, we get $x+x \frac{10}{x}+x \frac{100}{x^{2}}=35$, which simplifies to $x^{2}-25 x+100=0$. The solutions to this equation are $x_{1}=5$ and $x_{2}=20$. Finally, the solutions to the given system are $(5,2)$ and $\left(20, \frac{1}{2}\right)$.