<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] Since the expressions you know the value of are homogeneous and symmetric in $a,b,c$, we know (or should know) they are expressible in terms of homogeneous polynomials in the elementary symmetric functions $E_1=a+b+c$, $E_2=ab+ac+bc$, $E_3=abc$ and in polynomial terms.
In order to find these expressions I prefer a more systematic approach, which can be as follows, rather than plugging in the values and going by attempts.
Degree considerations tell us that
$$
a^2+b^2+c^2=\alpha E_1^2+\beta E_2
$$
With $(a,b,c)=(1,0,0)$ we obtain $1=\alpha$; with $(a,b,c)=(1,1,0)$ we get $2=4\alpha+\beta$. Thus $\beta=-2$.
Also
$$
a^3+b^3+c^3=\alpha E_1^3+\beta E_1E_2+\gamma E_3
$$
With $(a,b,c)=(1,0,0)$ we obtain $1=\alpha$; with $(a,b,c)=(1,1,0)$ we get $2=8\alpha+2\beta=0$; with $(a,b,c)=(1,1,1)$ we get $3=27\alpha+9\beta+\gamma$. Therefore $\beta=-3$ and $\gamma=3$.
The relations are then
$$
\begin{cases}
a^2+b^2+c^2=E_1^2-2E_2\\[4px]
a^3+b^3+c^3=E_1^3-3E_1E_2+3E_3
\end{cases}
$$
Since you're given $E_1=2$, $a^2+b^2+c^2=6$ and $a^3+b^3+c^3=8$, the relations become
$$
E_2=-1,\qquad
E_3=2E_2=-2
$$
Why is this important? Because given $E_1$, $E_2$ and $E_3$, the polynomial having $a,b,c$ as roots is
$$
x^3-E_1x^2+E_2x-E_3
$$
(Viète's formulas). Thus the polynomial is
$$
x^3-2x^2-x+2=x^2(x-2)-(x-2)=(x^2-1)(x-2)=(x-1)(x+2)(x-2)
$$
and you found the roots.
</s>