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 $a,b,c$?

$$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(a+c)$$
$$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+ab(2-c)+bc(2-a)+ca(2-b)$$
$$12=8+2(ab+bc+ca)-3abc$$
Now,
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
$$2(ab+bc+ca)=4-6=-2$$
Thus,
$$12=8-2-3abc$$
$$abc=-2$$
Let
$$f(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
$$f(x)=x^3-2x^2-x+2$$
The values of $a,b,c$ are the roots of $f(x)$.
By trial and error, clearly $1$ is a root.
Then, $$x^3-2x^2-x+2=(x-1)(x^2-x-2)=(x-1)(x+1)(x-2)$$
The values of $a,b,c$ are thus $-1,1,2$ in any order.
