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 $a,b,c$?

Note this convenient fact: $$(a+b+c)^3-(a^3+b^3+c^3) = 3(a+b)(b+c)(c+a)$$
Since we know the left-hand side is zero, we also know the right-hand side is zero; this tells us that at least one of these is true:
$$a+b=0 \qquad b+c=0 \qquad c+a=0 \tag{$\star$}$$
We'll go ahead and take the middle one (the others will work out similarly), so that $c = -b$. Then, we have
$$2 = a + b + c = a + 0 = a$$
which gives us $a$. Also
$$6 = a^2 + b^2 + c^2 = 2^2 + b^2 + (-b)^2 = 4 + 2 b^2$$
which gives us $b^2 = 1$, so that $b$ is either $+1$ or $-1$, and $c$ is its opposite.
If we'd made a different choice at $(\star)$, we'd have gotten the same values, $2$, $1$, $-1$, assigned to different variables. That's to be expected, since the original equations are symmetric in $a$, $b$, $c$; the letters are interchangeable. Therefore, the best we can say here is that 

$$\{ a, b, c\} = \{2, 1, -1\}$$

