The South African Mathematical Olympiad Third Round

By putting $X=x+2$, $Y=y+2$ and $Z=z+2$, the system of equations translates to

$$(X-1)(Y-1)(Z-1)=3\dots (1)$$

$$XYZ=-2\dots (2)$$

$$(X+1)(Y+1)(Z+1)=-1.\dots (3)$$

By expanding the left hand sides of (1) and (3), and substituting $XYZ=-2$, we obtain the system

$$\begin{array}{rl}{T}_1-T_2& =6\\ T_1+T_2& =0,\end{array}$$

where $T_1=X+Y+Z$ and $T_2=XY+YZ+ZX$. From this, we solve $T_1=3$ and $T_2=-3$, so that

$$\begin{array}{rl}(x+20)(y+20)(z+20)& =(X+18)(Y+18)(Z+18)\\ & =-2+18\cdot T_2+18^2\cdot T_1+18^3\\ & =-2+18\cdot (-3)+18^2\cdot (3)+18^3\\ & =6748.\end{array}$$