1. For which values of the parameter $a$ does the system

$$
\begin{aligned}
& x^{3}-a y^{3}=\frac{1}{2}(a+1)^{2} \\
& x^{3}+a x^{2} y+x y^{2}=1
\end{aligned}
$$

have at least one solution $(x, y)$ that satisfies the condition $x+y=0$?
Solution. Let's assume that the system (1) has a solution $(x, y)$ that satisfies the condition $x+y=0$; thus, $y=-x$, and substituting into both equations, we get the equations

$$
\begin{aligned}
& (1+a) x^{3}=\frac{1}{2}(a+1)^{2} \\
& (2-a) x^{3}=1
\end{aligned}
$$

From equation (2) we get $a=-1$ or $x^{3}=\frac{1}{2}(a+1)$, for $a \neq-1$. Substituting $a=-1$ into (3) gives $x=\frac{\sqrt[3]{9}}{3}$, and substituting $x^{3}=\frac{1}{2}(a+1)$ for $a \neq-1$ gives

$$
\frac{1}{2}(a+1)=\frac{1}{2-a} \text {, i.e., } a=0 \text { and } a=1 .
$$

Therefore, for $a=0, a=1$, and $a=-1$, the system (1) may have a solution of the form $(x,-x)$.

Now, let's see if for these values of $a$ the system (1) has a solution of that form.

For $a=0$, from the first equation of (1) we get $x=\frac{\sqrt[3]{4}}{2}$, and for this value of $x$, from the second equation of (1) we get $y= \pm \frac{\sqrt[3]{4}}{2}$. Thus, in this case, a solution to the system is the pair $\left(\frac{\sqrt[3]{4}}{2},-\frac{\sqrt[3]{4}}{2}\right)$.

For $a=1$, we get the system

$$
\left\{\begin{array} { l } 
{ x ^ { 3 } - y ^ { 3 } = 2 } \\
{ x ^ { 3 } + x ^ { 2 } y + x y ^ { 2 } = 1 }
\end{array} , \text { , i.e., } \quad \left\{\begin{array}{l}
(x-y)\left(x^{2}+x y+y^{2}\right)=2 \\
x\left(x^{2}+x y+y^{2}\right)=1
\end{array}\right.\right.
$$

one solution of which is the pair $(1,-1)$.

For $a=-1$, we get the system

$$
\left\{\begin{array}{l}
x^{3}+y^{3}=0 \\
x^{3}+x^{2} y+x y^{2}=1
\end{array}\right.
$$

one (solution) of which is $\left(\frac{\sqrt[3]{9}}{3},-\frac{\sqrt[3]{9}}{3}\right)$.

Consequently, the system (1) has a solution of the form $(x,-x)$ only for $a=0, -1, 1$.