Since Mr. Wee did Q1 and Q3, I shall try Q2
(a/bc)+(b/ca)+(c/ab)-(1/a)-(1/b)-(1/c)
= (a/bc) - (1/c) + (b/ca) - (1/a) +
(c/ab) - (1/b)
= (a-b)/bc + (b-c)/ac + (c-a)/ab
from a+x^2=2006, b+x^2=2007 and c+x^2=2008
a = 2006 - x^2, b = 2007 - x^2, c = 2008 - x^2
a - b = -1
b - c = -1
c - a = 2
c - b = 1
Also, abc is given as 3
So, (a-b)/bc + (b-c)/ac + (c-a)/ab
= -1/bc - 1/ac + 2/ab
= (-a -b +2c)/abc
= ([c - a] + [c - b]) / abc
= (2 + 1) / 3
= 1 (answer)
Alternatively,
(a/bc)+(b/ca)+(c/ab)-(1/a)-(1/b)-(1/c) = (a^2 + b^2 + c^2 - ab -
bc - ac)/3 =(-a - b + 2c)/3 = 3/3 = 1