Solve the following system of equations:

$$
\begin{array}{r}
x^{3}+y^{3}=400 \\
x^{2} y+x y^{2}=200
\end{array}
$$
Solution. If we add three times the second equation to the first equation,

$$
x^{3}+3 x^{2} y+3 x y^{2}+y^{3}=1000
$$

results. According to a notable identity, the expression on the left side is equal to $(x+y)^{3}$. Taking the cube root of both sides:

$$
x+y=10
$$

Expressing the value of $y$ and substituting it into the first equation, after simplification, we get:

$$
x^{2}-10 x+20=0
$$

The roots of this equation are: $x_{1}=5+\sqrt{5} ; x_{2}=5-\sqrt{5}$. Since the system of equations is symmetric in $x$ and $y$, the solutions are:

$$
\begin{aligned}
& x_{1}=5+\sqrt{5} \text { and } x_{2}=5-\sqrt{5} \\
& y_{1}=5-\sqrt{5} \text { and } y_{2}=5+\sqrt{5}
\end{aligned}
$$

By substitution, we can verify the correctness of the solutions.

Patkós Petra (Kunszentmárton, József Attila Gymn., III. grade)