\section*{Problem 3A - 261043A}

Determine all triples \(\left(x_{1}, x_{2}, x_{3}\right)\) of real numbers \(x_{1}, x_{2}, x_{3}\) that satisfy the following system of equations (1), (2), (3)!

\[
\begin{aligned}
x_{1}+x_{2}+x_{3} & =3 \\
x_{1}^{3}+x_{2}^{3}+x_{3}^{3} & =3 \\
x_{1} \cdot x_{2} \cdot x_{3} & =1
\end{aligned}
\]
I. If a triple \(\left(x_{1}, x_{2}, x_{3}\right)\) of real numbers satisfies the system of equations, then: Substituting from (1) \(x_{3}=3-x_{1}-x_{2} \quad(4)\) into (2), we get

\[
x_{1}^{3}+x_{2}^{3}+27-27\left(x_{1}+x_{2}\right)+9\left(x_{1}+x_{2}\right)^{2}-x_{1}^{3}-3 x_{1}^{2} x_{2}-3 x_{1} x_{2}^{2}-x_{2}^{3}=3
\]

after dividing by 3, this becomes

\[
9\left(x_{1}+x_{2}\right)-3\left(x_{1}+x_{2}\right)^{2}+x_{1} x_{2}\left(x_{1}+x_{2}\right)=8
\]

Substituting (4) into (3), we get

\[
3 x_{1} x_{2}-x_{1} x_{2}\left(x_{1}+x_{2}\right)=1
\]

For the two numbers

\[
p=x_{1} x_{2} \quad(7) \quad ; \quad s=x_{1}+x_{2}
\]

equations (5) and (6) become

\[
9 s-3 s^{2}+p s=8 \quad(9) \quad ; \quad 3 p-p s=1
\]

After adding and then dividing by 3, we get

\[
p=s^{2}-3 s+3 \quad \Rightarrow \quad p=s^{2}-3 s+3
\]

Substituting into (9) yields

\[
9 s-3 s^{2}+s^{3}-3 s^{2}+3 s=8 \quad \Rightarrow \quad(s-2)^{3}=0 \quad \Rightarrow \quad s=2
\]

From (11), (7), and (8), we get

\[
p=1 \quad ; \quad x_{1} x_{2}=1 \quad ; \quad x_{1}+x_{2}=2
\]

Substituting \(x_{2}=2-x_{1}\) from (13) into (12), we get \(x_{1}=1\) and thus from (3) and (4) \(x_{2}=1, x_{3}=1\). Therefore, only the triple (1,1,1) can satisfy the system of equations (1), (2), (3), which is confirmed by verification.

\section*{Adapted from \([5]\)}