Solve the following system of equations:

$$
\begin{aligned}
& \frac{1}{x}+\frac{1}{2 y}=\left(x^{2}+3 y^{2}\right)\left(3 x^{2}+y^{2}\right) \\
& \frac{1}{x}-\frac{1}{2 y}=2\left(y^{4}-x^{4}\right)
\end{aligned}
$$
Solution. Let's perform the indicated operations on the right-hand sides of the equations:

$$
\begin{aligned}
& \frac{1}{x}+\frac{1}{2 y}=3 x^{4}+10 x^{2} y^{2}+3 y^{4} \\
& \frac{1}{x}-\frac{1}{2 y}=2 y^{4}-2 x^{4}
\end{aligned}
$$

We multiply the sum of the two equations by \( x \), and the difference by \( y \). (Clearly, \( x \) and \( y \) are not 0.) Thus, we get

$$
\begin{aligned}
& 2=x^{5}+10 x^{3} y^{2}+5 x y^{4} \\
& 1=5 x^{4} y+10 x^{2} y^{3}+y^{5}
\end{aligned}
$$

Notice that the right-hand sides of the equations we now have are precisely the terms of the binomial expansion of

$$
(x+y)^{5}=x^{5}+5 x^{4} y+10 x^{3} y^{2}+10 x^{2} y^{3}+5 x y^{4}+y^{5}
$$

In equation (3), the terms on the right-hand side are those where the exponent of \( y \) is even, and in equation (4), the terms where the exponent of \( y \) is odd. The sum of the two right-hand sides is therefore \((x+y)^{5}\), and their difference is \((x-y)^{5}\):

$$
3=(x+y)^{5}, \quad \text { i.e., } \quad \sqrt[5]{3}=x+y \quad \text { and } \quad 1=(x-y)^{5}, \quad \text { i.e., } \quad 1=x-y
$$

From here, the solution to the system of equations is:

$$
x=\frac{\sqrt[5]{3}+1}{2} \quad \text { and } \quad y=\frac{\sqrt[5]{3}-1}{2}
$$