Find the value of $\frac{(x+1)(y+1)}{x+y}+\frac{(x+1)(z+1)}{x+z}+\frac{(y+1)(z+1)}{y+z}$

Question

Find the value of $\frac{(x+1)(y+1)}{x+y}+\frac{(x+1)(z+1)}{x+z}+\frac{(y+1)(z+1)}{y+z}$ given that $x+y,$ $x+z$ and $y+z$ are distinct from $0$, $\;x+y+z=3$, and $xy+xz+yz=-1$.

Answer 1

Since $xy=-1-xz-yz=-1-z(x+y)$, we obtain $$\frac{(x+1)(y+1)}{x+y}=\frac{xy+x+y+1}{x+y}=\frac{(-1-z(x+y))+x+y+1}{x+y}=-z+1$$