<s>[INST] Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
 [/INST] You may use Newton's formulas for recovering the values of $ab+ac+bc$ and $abc$, in order to compute $a,b,c$ from the factorization of $(x-a)(x-b)(x-c)$. As an alternative, we may understand which conditions on $P_1,P_2,P_3$ grant that if $a+b=P_1$ and $a^2+b^2=P_2$, then $a^3+b^3=P_3$. We have:
$$ P_1 P_2 = P_3 + ab(a+b) = P_3+P_1\cdot\frac{P_1^2-P_2}{2} $$
or:
$$ 2 P_3 = 3 P_1 P_2 - P_1^3.$$
Assuming $c=1$, we have $(P_1,P_2,P_3)=(1,5,7)$ that fulfills the previous condition, hence we may assume $a+b=1$ and $a^2+b^2=5$. The solutions are so given by the permutations of $\color{red}{(-1,1,2)}$.
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