Simple System of Equations Part 3

Expanding $(x+1)(y+1)(z+1) = xyz + xy + yz + zx + x + y + z +1$

We note that adding the three given equations we have:

$\begin{cases} xyz + 3x + 3yz = 106 \\ xyz + 3z + 3xy = 79 \\ xyz + 3y + 3zx = 82 \end{cases} \Rightarrow 3(xyz + xy + yz + zx + x + y + z) = 267$

$\Rightarrow (x+1)(y+1)(z+1) = \dfrac{267}{3} +1 = 89 + 1 = \boxed{90}$

we have:A= (x+1)(y+1)(z+1)=xyz+x+y+z+xy+yz+zx+1, so when we plus both 3 equations, we have 3(xyz+x+y+z+xy+yz+zx)=267 <=> 3(A-1)=267, so A=90