414. Solve the system of equations:

$$
\left\{\begin{array}{l}
x \sqrt{y}+y \sqrt{x}=6 \\
x^{2} y+y^{2} x=20
\end{array}\right.
$$
$\triangle$ Let's introduce two new variables:

$$
x=\sqrt{y}, \quad y \sqrt{x}=t
$$

where $z \geq 0, t \geq 0$. Then

$$
z+t=6, \quad z^{2}+t^{2}=20
$$

The solutions of the last system are $(4 ; 2)$ and $(2 ; 4)$. Now let's consider two cases.

1) For the first solution of this system, we get:

$$
x \sqrt{y}=4, \quad y \sqrt{x}=2
$$

from which

$$
x^{2} y=16, \quad y^{2} x=4
$$

Multiply and divide the equations of the last system:

$$
x^{3} y^{3}=64, \quad \frac{x}{y}=4
$$

Then

$$
x y=4, \quad x=4 y
$$

Therefore, we will have:

$$
4 y^{2}=4, \quad y^{2}=1 ; \quad y=1, \quad x=4
$$

(The value $y=-1$ is not suitable).

2) Similarly, in the second case, we get the solution $(1 ; 4)$.

Answer: $(4 ; 1),(1 ; 4)$.