<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] One possibility to determine this---which is by no means perfect, and will not work all of the time!---is to try a few small integers. The best one to focus on is the second equation:
$$
a^2 + b^2 + c^2 = 6
$$
since all of the terms must be positive. Note that if $a = \pm3$, say, then $a^2 = 9$, which means that all terms have to be from the set $\{-2, -1, 0, 1, 2\}$. They can't all be $\pm1$, since that would be to small, so at least one has to be $\pm2$.
Can you see where to go from here? You can also simplify by noting that the solution is symmetric under swapping $a \to b \to c$ (or any other permutation), so you can assume for simplicity that $a$ is the largest in absolute value.
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