<s>[INST] Determine $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$ $(x_{i},y_{i}), i=1,2,3$ are solutions for
$$ x^{3} - 3xy^{2} = 2010 $$
$$ y^{3} - 3yx^{2} = 2009 $$
What is $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$?

Attempt:
My approach is that we should look for $\frac{y-x}{y}=$something.
Notice that
$$ x^{3} - y^{3} + 3(yx^{2} - xy^{2}) = 1  $$
$$ (x-y)(x^{2} + y^{2} + xy) + 3xy(x-y) = 1 $$
$$ (y-x)(x^{2} + y^{2} + 4xy)  = -1$$
$$ \frac{y-x}{y} = -\frac{1}{y(x^{2} + y^{2} + 4xy)} (\text{does not seemt to go anywhere})$$
Anther approach if I sum the equations:
$$ x^{3} + y^{3} - 3(xy^{2} + yx^{2}) = 4019 $$
$$ (x + y)(x^{2} + y^{2} - xy) - 3(xy)(x + y) = 4019 $$
$$ (x+y)(x^{2} + y^{2} - 4xy) = 4019 $$
Multiply both we get
$$ (x^{2}-y^{2})((x^{2}+y^{2})^{2} -16 (xy)^{2}) = 4019 $$
???
 [/INST] We have $xy\ne0$
Let $x=my$
$$\dfrac{2010}{2009}=\dfrac{m^3y^3-3(my)y^2}{y^3-3(my)^2y}=\dfrac{m^3-3m}{1-3m^2}$$
Now set $1-m=n\iff m=1-n$
Replace $m$ with $1-n$ to form a cubic equation in $n$
Now use Vieta's formula
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