Finding all the sets of three real numbers that satisfy specific equations I am having trouble finding a set of three numbers of real numbers $(x, y, z)$ satisfying $x + y + z = xy + xz + yz = 3$. I have tried factoring the equations around but I'm not having any luck and I don't think I'm going on the right path.

Clearly $a=b=c=1$ is a solution.
Let $a=(1-\alpha)$, and $b=(1-\beta)$. Then $a+b+c=3$ implies 
$c = (1 + \alpha + \beta)$
\begin{align}
   ab + bc + ca &= 3 \\
   (1-\alpha)(1-\beta) + 
   (1-\beta)(1 + \alpha + \beta) + 
   (1 + \alpha + \beta)(1-\alpha) &= 3 \\
   -\alpha^2 - \alpha \beta - \beta^2 + 3 &= 3 \\
   \alpha^2 + \alpha \beta + \beta^2 &= 0 \\
   2\alpha^2 + 2\alpha \beta + 2\beta^2 &= 0 \\
   \alpha^2 + \beta^2 + (\alpha^2+2\alpha \beta + \beta^2) &= 0 \\
   \alpha^2 + \beta^2 + (\alpha + \beta)^2 &= 0
\end{align}
Which implies $\alpha = \beta = 0$. So $a=b=c=1$ is the only solution.
