Solve the following system of equations:

$$
\begin{gathered}
\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}=3 \ldots \\
x^{3}+y^{3}=2 \ldots
\end{gathered}
$$
1)-ból $\quad \frac{(x+y)^{2}-x y}{(x+y)^{2}-3 x y}=3$.

After removing the fraction and rearranging:

$$
(x+y)^{2}=4 x y \ldots
$$

which means

$$
(x-y)^{2}=0 \text{, so } x=y \text{. }
$$

2)-by substituting

$$
2 x^{3}=2, \quad x=1=y \text{. }
$$

Winter Miksa (Kegyesrendi g. V. o. Szeged.)[^0]


[^0]:    $11 x^{2}-x y+y^{2} \neq 0$. Indeed, $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=2$.

    $x^{2}-x y+y^{2}=0 \quad$ leads to a contradiction $\quad(x+y) \cdot 0=2$.

    22 We are focusing on real values.