6. Find all possible values of the sum $x+y$, where real numbers $x$, $y$ satisfy the equation $x^{3}+y^{3}=3 x y$.
SOLUTION. We are actually looking for all values of the parameter $s$ for which the system of equations

$$
\begin{aligned}
x^{3}+y^{3} & =3 x y \\
x+y & =s
\end{aligned}
$$

has solutions in the set of real numbers. From the second equation, we express $y=s-x$ and substitute it into the first equation, which we will solve with respect to the unknown $x$:

$$
\begin{gathered}
x^{3}+(s-x)^{3}=3 x(s-x), \\
x^{3}+s^{3}-3 s^{2} x+3 s x^{2}-x^{3}=3 s x-3 x^{2} \\
3(s+1) x^{2}-3 s(s+1) x+s^{3}=0 .
\end{gathered}
$$

This equation clearly has no solution for $s=-1$. For $s \neq-1$, it is a quadratic equation which has solutions in the set of real numbers if and only if its discriminant $D$ is non-negative. By calculation,

$$
D=9 s^{2}(s+1)^{2}-12(s+1) s^{3}=3 s^{2}(s+1)(3-s)
$$

we find that $D \geqq 0$ if and only if $-1 \leqq s \leqq 3$, which together with the condition $s \neq-1$ gives the set of possible values of the sums $s$: it is the half-open interval $(-1,3)$.

In retrospect, it can be seen that for each $s$ from the interval $(-1,3\rangle$ there exists a number $x$ which is a root of the above quadratic equation, and that these numbers $x$ and the corresponding value $y=s-x$ satisfy the equation $x^{3}+y^{3}=3 x y$.

For completeness, we calculate the numbers $x, y$ which are solutions to the considered system for any $s \in(-1,3\rangle$:

$$
x_{1,2}=\frac{3 s(s+1) \pm s \sqrt{3(s+1)(3-s)}}{6(s+1)}=\frac{s}{2} \pm s \sqrt{\frac{3-s}{12(s+1)}} .
$$

By substituting into $y=s-x$, we find that for each $s$ there are two pairs

or

$$
[x, y]=\left\{\frac{s}{2}+s \sqrt{\frac{3-s}{12(s+1)}}, \frac{s}{2}-s \sqrt{\frac{3-s}{12(s+1)}}\right\}
$$

$$
[x, y]=\left\{\frac{s}{2}-s \sqrt{\frac{3-s}{12(s+1)}}, \frac{s}{2}+s \sqrt{\frac{3-s}{12(s+1)}}\right\} .
$$

The equality $x^{3}+y^{3}=3 x y$ can be verified for the found $x$ and $y$ either by substitution and direct calculation, or using Vieta's formulas for the roots of the above quadratic equation

$$
x+y=s, \quad x y=\frac{s^{3}}{3(s+1)}
$$

according to which

$$
x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)=s^{3}-3 s \frac{s^{3}}{3(s+1)}=\frac{s^{3}}{s+1}=3 x y .
$$

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