Problem
ALG-B1-M06-P018 Recover the triple
#18
★★★★☆ Level 4 of 5
Let \(a+b+c=0\), \(a^2+b^2+c^2=6\), \(a^3+b^3+c^3=6\). Find all possible triples \((a,b,c)\).
Find \(ab+bc+ca\) and \(abc\), then form the cubic polynomial.
From the first two conditions, \(ab+bc+ca=-3\). From \(a^3+b^3+c^3=3abc\) we get \(abc=2\).
Thus \(a,b,c\) are roots of \(t^3-3t-2=0\). The factorization \(t^3-3t-2=(t-2)(t+1)^2\) gives the roots \(2,-1,-1\). Hence all solutions are permutations of \((2,-1,-1)\).
The problem reinforces the transition from symmetric sums to a polynomial.