## Task $9 / 71$

For which real numbers $a$ does the system of equations

$$
\begin{array}{r}
2 x^{3}-2 a y^{3}=(a+1)^{2} \\
x^{3}+a x^{2} y+x y^{2}=1
\end{array}
$$

have exactly one real solution $(x ; y)$ with $x=-y$?
By the requirement $x=-y$, the system transforms into the equations

$$
2 x^{3}(a+1)=(a+1)^{2} \quad ; \quad x^{3}(2-a)=1
$$

The first equation is certainly satisfied for $a=-1$. It then follows from the second equation that the solution is $x=-y=\sqrt[3]{\frac{1}{2}}$.

For $a \neq 1$, the first equation transforms into $2 x^{3}=a+1 \quad ; \quad x^{3}=\frac{a+1}{2}$.

Substituting into the second equation, after a brief calculation, we obtain the quadratic equation in $a$: $a^{2}-a=0$ with solutions $a=0$ and $a=1$. The corresponding solutions are $x=-y=\sqrt[3]{\frac{1}{2}}$ and $x=-y=1$.

Thus, exactly one solution $(x ; y=-x)$ is obtained for the values $a_{1}=-1, a_{2}=0$, and $a_{3}=+1$.