3. Let $x, y$, and $z$ be real numbers that satisfy the following system:
$$
\begin{aligned}
x+3 y+6 z & =1 \\
x y+2 x z+6 y z & =-8 \\
x y z & =2
\end{aligned}
$$

Find the smallest possible value of $x+y+z$.
Answer: $-\frac{8}{3}$
Solution: Let $y^{\prime}=3 y, z^{\prime}=6 z$. The system of equations can then be rewritten as
$$
\begin{aligned}
x+y^{\prime}+z^{\prime} & =1 \\
x y^{\prime}+x z^{\prime}+y^{\prime} z^{\prime} & =-24 \\
x y^{\prime} z^{\prime} & =36
\end{aligned}
$$
and so $x, y^{\prime}, z^{\prime}$ are the roots of the polynomial $t^{3}-t^{2}-24 t-36=0$. This factors into $(t-6)(t+$ $3)(t+2)=0$, and so, in some order, $x, y^{\prime}, z^{\prime}$ must be $6,-3,-2$. This gives us the following solutions: $\left(6,-1,-\frac{1}{3}\right),\left(6,-\frac{2}{3},-\frac{1}{2}\right),\left(-3,2,-\frac{1}{3}\right),\left(-3,-\frac{2}{3}, 1\right),(-2,-1,1),\left(-2,2,-\frac{1}{2}\right)$. Among these, the ordered triple which gives the minimum sum is $\left(-3,-\frac{2}{3}, 1\right)$, which has a sum of $-8 / 3$.