Polynomial question !

( b-1 )( a-1 ) = ab - a - b + 1


f( x ) = ( x-b ) ( x-a ) ( x-1 ) = 0

f( 2 ) = ( 2-b ) ( 2-a ) ( 2-1 ) = 0
4 - 2a - 2b + ab = 0.........................(1)

f( 3 ) = ( 3-b ) ( 3-a ) ( 3-1 ) = 0
( 9 - 3a - 3b + ab ) ( 2 ) = 0
9 - 3a - 3b + ab =0.........................(2)

(1)-(2) -5 + a + b = 0
a + b = 5

sub a + b = 5 into (1)
4 - 2a - 2b + ab = 0
4 - 2( a + b ) + ab = 0
4 - 2(5) + ab = 0
ab = 6


consider
( b-1 )( a-1 )
= ab - a - b + 1
= ab - (a + b) + 1
= 6 - 5 + 1
=2

Remarks: you can put any real no into f(x) to solve this question(except the root of f(x) , ie : a, b and 1). In this case, I choosed x = 2 and x = 3

Since f(x) = (x-b) (x-a) (x-1) = 0

f(0) = -ab = 0 --------(1)

f(-1) = -2 (b+1) (a+1) = 0
=> ab + a + b + 1 = 0
=> a + b = -1 - ab
=> a + b = -1 ------------(2)

(b-1) (a-1) = ab - a - b + 1
= 0 - (a+b) +1
= 0 - (-1) + 1
= 2