If $x^3 + 2x^2 -ax - b$ is exactly divisible by $(x^2 - 1)$, then the values of a and b are :

a = 1 and b = 2

If $x^3 + 2x^2 -ax - b$ is exactly divisible by $(x^2 - 1)$

 So, (x² - 1) = 0 

= x = 1, -1 

Put x = 1

= 1³ + 2(1)² - a - b = 0 

= a + b = 3 ----(a)

Put x = -1

= (-1)³ + 2(-1)² + a - b = 0

= a - b = -1 ----(b)

 = from eq(a) and (b) 

= a = 1, b = 2