Let's find the solutions of the following system of equations in the set of non-negative numbers:

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

Do we get other solutions if we only assume that $x+y>0$?
I. solution. If $x \geq 0$ and $y \geq 0$, then from (1) it follows that $x \leq 1$ and $y \leq 1$, and from (2) it follows that $x \geq 1$. Thus, only $x=1$ can hold. Substituting this into (2) yields $y=0$. Substituting these values shows that this pair of numbers is also a solution to (1). Therefore, the solution to the system of equations in the realm of non-negative numbers is: $x=1, y=0$.

If $x+y>0$, then according to $(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$,

$$
(x+y)^{2}=\frac{(x+y)^{3}}{x+y}=\frac{x^{3}+y^{3}}{x+y}+3 x y
$$

If $x+y>1$, then the left side is greater than 1, and the value of the fraction on the right side increases if the denominator is reduced to 1. We thus get that under the condition $10$, $x+y=1$ follows. Substituting this into (2) gives us

$$
x^{2}-y^{2}=(x+y)(x-y)=x-y=1
$$

Therefore, the system of equations under the condition $x+y>0$ is equivalent to the system

$$
x+y=1, \quad x-y=1
$$

and this still has only the solution $x=1, y=0$.

II. solution. We will show that the system of equations has no other solution if we do not impose any conditions on $x$ and $y$.

The first equation can be factored using the identities $x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$ and $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$:

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

The second factor can be expressed as a sum of squares:

$$
x^{2}+y^{2}-x y+x+y+1=\frac{1}{2}\left[(x+1)^{2}+(y+1)^{2}+(x-y)^{2}\right]
$$

Thus, equation (1) is equivalent to the equation

$$
(x+y-1)\left[(x+1)^{2}+(y+1)^{2}+(x-y)^{2}\right]=0
$$

From this, either $x+y=1$ or $x=y=-1$.

The second case is not a solution due to (2), and in the first case, from (2) we get the solution $x=1$ and $y=0$.