, 2015

Solution:

Consider the cubic polynomial $f(t) = (x + t)(y + t)(z + t)$. By the theory of finite differences, $f(3) - 3 f(2) + 3 f(1) - f(0) = 3! = 6$, since $f$ is monic. Thus
$$f(3) = 6 + 3 f(2) - 3 f(1) + f(0) = 6 + 3(-3) - 3(7) + (-4) = -28.$$

Solution 2:

Alternatively, note that the system of equations is a (triangular) linear system in $w := x y z$, $v := x y + y z + z x$, and $u := x + y + z$. The unique solution $(u, v, w)$ to this system is $\left(-\frac{27}{2}, \frac{47}{2}, -4\right)$. Plugging in yields
$$\begin{aligned} (x+3)(y+3)(z+3) & = w + 3v + 9u + 27 \\ & = -4 + 3 \cdot \frac{47}{2} + 9 \cdot \left(-\frac{27}{2}\right) + 27 \\ & = -28 \end{aligned}$$