If α and β are the roots of the equation x2-a(x+1)-b=0 then (α+1)(β+1)=
b
-b
1-b
b-1
Given equation x2-a(x+1)-b=0
⇒x2-ax-a-b=0⇒α+β=a, αβ=-(a+b)
Now (α+1)(β+1)=αβ+α+β+1
=-(a+b)+a+1=1-b