Solve the following system of equations:

$$
2 \sqrt{\sqrt{x}+\sqrt{y}}+\sqrt{x}+\sqrt{y}=8, \quad \sqrt{x^{3}}+\sqrt{y^{3}}=40
$$
$$
\text { (1) } 2 \sqrt{\sqrt{x}+\sqrt{y}}+\sqrt{x}+\sqrt{y}=8, \quad \text { (2) } \quad \sqrt{x^{3}}+\sqrt{y^{3}}=40
$$

Equation (1) in terms of the variable $z=\sqrt{\sqrt{x}+\sqrt{y}}$ represents the equation

$$
2 z+z^{2}=8
$$

whose roots are $z_{1}=-4, z_{2}=2$. In our case, the first root cannot be considered because $\sqrt{\sqrt{x}+\sqrt{y}}$ is non-negative. Therefore, (1) is equivalent to the equation

$$
\sqrt{x}+\sqrt{y}=4
$$

Substituting the expression $\sqrt{y}=4-\sqrt{x}$ obtained from this into (2) in place of $\sqrt{y}$, and performing the indicated operations, we get the equation

$$
x-4 \sqrt{x}+2=0
$$

From this, $2 \pm \sqrt{2}$ is obtained for $\sqrt{x}$, and since these are positive, and the values obtained for $\sqrt{y}$ from them based on (3) are also positive, the solutions of the (1)-(2) system are:

From the value $\sqrt{x}=2+\sqrt{2}$, we get $x_{1}=6+4 \sqrt{2}, \quad y_{1}=6-4 \sqrt{2}$;

From the value $\sqrt{x}=2-\sqrt{2}$, we get $x_{1}=6-4 \sqrt{2}, \quad y_{2}=6+4 \sqrt{2}$.