Solving system of equation in two variables If $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ are the real solutions of two equation $$x^3-3xy^2=2005$$ and $$y^3-3x^2y=2004$$ then find value of 
$$\frac{y_1 y_2 y_3}{2(y_1-x_1)(y_2-x_2)(y_3-x_3)}.$$
I added and subtracted the two equation to get $$(x-y)(x^2+4xy+y^2)=1$$ and $$(x+y)(x^2-4xy+y^2)=4009$$ but couldn't proceed further. Then I used vieta's formulas in second equation  to get $y_1 y_2 y_3=2004$ but couldn't resolve the denominator similarly. Now I am stuck please help me out. 

Hint:  by brute force, let $t=x/y\,$, then:
$$
\begin{align}
t^3 - 3 t = \frac{2005}{y^3} \\[5px]
-3t^2 + 1 = \frac{2004}{y^3}
\end{align}
$$
Dividing the two:
$$
\frac{t^3-3t}{-3t^2+1} = \frac{2005}{2004} \quad\iff\quad 2004 t^3 + 6015 t^2 -6012 t - 2005 = 0
$$
Now $\;\displaystyle \frac{{y_1 \cdot y_2 \cdot y_3}}{{2(y_1-x_1)(y_2-x_2)(y_3-x_3)}} = \frac{1}{2(1-t_1)(1-t_2)(1-t_3)}\,$ can be calculated using Vieta's.
