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Exercise 2135Algebra › Equations and radicalsOlympiadAdded 3 Nov 2026

A symmetric 3-variable system

#symmetric system#Vieta#olympiad

Statement

Solve in R3\mathbb{R}^3 the system
{x+y+z=6xy+yz+zx=11xyz=6\begin{cases}x+y+z=6\\ xy+yz+zx=11\\ xyz=6\end{cases}

Hint

x,y,zx,y,z are the roots of t3−6t2+11t−6=0t^3-6t^2+11t-6=0.

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Solution

Polynomial. By Vieta's formulas, x,y,zx,y,z are roots of
t3−St2+Qt−P=0t^3-St^2+Qt-P=0
with S=6S=6, Q=11Q=11, P=6P=6:
t3−6t2+11t−6=0.t^3-6t^2+11t-6=0.

Factor. t=1t=1 is a root: 1−6+11−6=01-6+11-6=0. Divide: t3−6t2+11t−6=(t−1)(t2−5t+6)=(t−1)(t−2)(t−3)t^3-6t^2+11t-6=(t-1)(t^2-5t+6)=(t-1)(t-2)(t-3).

Roots. {x,y,z}={1,2,3}\{x,y,z\}=\{1,2,3\} (in any order).

Conclusion. The solutions are the six permutations of (1,2,3)(1,2,3).