Determine $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$

Question

$(x_{i},y_{i}), i=1,2,3$ are solutions for

$$ x^{3} - 3xy^{2} = 2010 $$

$$ y^{3} - 3yx^{2} = 2009 $$

What is $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$?

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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

$$ (x^{2}-y^{2})((x^{2}+y^{2})^{2} -16 (xy)^{2}) = 4019 $$

???

Answer 1

It is easy to see none of the $y_i$ vanishes.

Let $u = \frac{x}{y}$ and $u_i = \frac{x_i}{y_i}$. $u_i$ are roots of the equation:

$$x^3 - 3xy^2 - \frac{2010}{2009}(y^3 - 3yx^2) = 0 \iff u^3 - 3u - \frac{2010}{2009}(1 - 3u^2) = 0 $$ This implies $$(u-u_1)(u-u_2)(u - u_3) = u^3 - 3u - \frac{2010}{2009}(1 - 3u^2)$$ The value we want is the value of this expression at $u = 1$.

$$\prod_{i=1}^3\left(1 - \frac{x_i}{y_i}\right) = \prod_{i=1}^3 (1 - u_i) = 1 - 3 - \frac{2010}{2009}(1-3) = \frac{2}{2009}$$

Answer 2

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