The complex numbers $x, y, z$ satisfy \[ \begin{aligned} x y z & =-4 \\ (x+1)(y+1)(z+1) & =7 \\ (x+2)(y+2)(z+2) & =-3 \end{aligned} \] Find, with proof, the value of $(x+3)(y+3)(z+3)$.
Solution 1. 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 \[ \begin{aligned} (x+3)(y+3)(z+3) & =w+3 v+9 u+27 \\ & =-4+3 \cdot \frac{47}{2}+9 \cdot\left(-\frac{27}{2}\right)+27 \\ & =-28 \end{aligned} \] Remark. Since $f(0)<0, f(1)>0, f(2)<0, f(+\infty)>0$, the intermediate value theorem tells us the roots $-x,-y,-z$ of $f$ are real numbers in $(0,1),(1,2)$, and $(2,+\infty)$, in some order. With a little more calculation, one finds that $x, y, z$ are the three distinct zeroes of the polynomial $X^{3}+\frac{27}{2} X^{2}+\frac{47}{2} X+4$. The three zeroes are approximately $-11.484,-1.825$, and -0.191 . $\fbox{-28}$.