maths problem

consider a useful expression that is easy to memorize
(a+b+c)^2 = a^2 +b^2 +c^2 +2(ab +bc +ca)

hence:
a^2 +b^2 +c^2 =(a+b+c)^2 -2(ab+bc+ca)

let a= 1^2, b=2007^2, c=2008^2
then
1^4 +2007^4 +2008^4
= (1^2)^2+ (2007^2)^2 + (2008^2)^2
= a^2 +b^2 +c^2
= (a+b+c)^2 -2(ab+bc+ca)
= [1^2 +2007^2 +2008^2]^2 - 2[(2007^2+(2007^2)x(2008^2)+2008^2]

hence
1^4 + 2007^4 + 2008^4 over
1^2 + 2007^2 + 2008^2

= (1^2 + 2007^2 + 2008^2)
- 2[(2007^2+(2007^2)x(2008^2)+2008^2] over
(1^2 + 2007^2 + 2008^2)

= (1+ 4028049 +4032064)
-2 (4028049 + 4028049 x4032064 +4032064] over
(1+ 4028049 +4032064)

= 8060114 - 2 ( 2015028.5)
(can't shown the numurator and the denominator)
= 4030057

(1^4 + 2007^4 + 2008^4)/(1^2 + 2007^2 + 2008^2)
=(1 + 16225178746401 + 16257540100096 )/(1 + 4028049 + 4032064 )
=32482718846498 / 8060114
=4030057