2015/014) if $a^3-3a^2b = 2005$ and $b^3-3b^2a = 2004$ then find

$(\dfrac{b_3-a_3}{b_3})(\dfrac{b_2-a_2}{b_2})(\dfrac{b_1-a_1}{b_1})$ 

We have

$a^3-3a^2b = 2005\cdots(1)$
and


$b^3-3b^2a = 2004\cdots(2)$

now subtract 2nd equation from 1st

$a^3-3a^2b+3ab^2-b^3= 1$
or $(a-b)^3=1$

$(a_1-b_1)\,(a_2-b_3)\,(a_3-b_3)$ are roots of equation $x^3-1=0$

so $(a_1-b_1)(a_2-b_3)(a_3-b_3) = 1\cdots(3)$

further $a=b+1$ and putting in (2) we get
$b^3-3b^2(b+1)=2004$
or   $2b^3 +3b^2= -2004$
so product of roots = $b_1b_2b_3=- 1002\cdots(4)$

from (3) and (4)

$(\dfrac{b_3-a_3}{b_3})(\dfrac{b_2-a_2}{b_2})(\dfrac{b_1-a_1}{b_1})= \dfrac{1}{1002}$

Note:

I have taken the problem in the site at  http://mathhelpboards.com/challenge-questions-puzzles-28/evaluate-value-product-14240.html#post67603

Where you can find some other different correct solutions.

Unforunately the problem stated in the link was wrong and I corrected the same