What relationship exists between the numbers $a, b, c$, if there is a pair of numbers $x, y$ for which the following equalities hold:
(1) $x+y=a$,
(2) $x^{2}+y^{2}=b$,
(3) $x^{3}+y^{3}=c$.
I. solution. The sum $x^{3}+y^{3}$ can be expressed using $x+y$ and $x^{2}+y^{2}$, based on the following identities:

$$
\begin{gathered}
x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right) \text { and } \\
x y=\left((x+y)^{2}-\left(x^{2}+y^{2}\right)\right)
\end{gathered}
$$

If $x$ and $y$ are values for which (1)-(3) hold, then by substitution and rearrangement, we find that the following relationship must hold between $a, b, c$:

$$
a^{3}-3 a b+2 c=0
$$

István Faragó (Budapest, Kálmán Könyves g. II. o. t.)

II. solution. The product of (1) and (2), with appropriate grouping:

$$
a b=(x+y)\left(x^{2}+y^{2}\right)=\left(x^{3}+y^{3}\right)+x y(x+y)
$$

We also take the following identity:

$$
a^{3}=(x+y)^{3}=\left(x^{3}+y^{3}\right)+3 x y(x+y)
$$

Subtracting three times (5) from this, the $x y$ term cancels out:

$$
a^{3}-3 a b=-2\left(x^{3}+y^{3}\right)=-2 c
$$

and this is identical to (4).

Györgyi Pap (Budapest, Ferenc Kölcsey g. II. o. t.)

III. solution. Besides the above clever manipulations, the usual elimination also leads to the goal. Substituting $y=a-x$ into (2) and (3):

$$
\begin{array}{ll}
x^{2}+(a-x)^{2}-b=0, & x^{3}+(a-x)^{3}-c=0 \\
2 x(x-a)+a^{2}-b=0 ; & 3 a x(x-a)+a^{3}-c=0
\end{array}
$$

Finally, subtracting twice the right-hand equality from the left-hand equality multiplied by $3 a$, we again obtain (4).

Remark. The essence of the question is that, considering (1)-(3) as a system of equations in $x, y$, we have 3 equations, which are only non-contradictory if the found (4) condition is satisfied.