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I found this question from my friend's math competition, I don't know where I must start it

There are 3 couples of real numbers $$(x_1,y_1) (x_2,y_2)$$ and $$(x_3, y_3)$$ that satisfies the system of equation : $$x³-3xy²=2010 , y³-3x²y=2009$$ Find the value of $$\left(1-\frac{x_1}{y_1}\right)\left(1-\frac{x_2}{y_2}\right)\left(1-\frac{x_3}{y_3}\right)$$

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Hint.

In the homogeneous polynomial

$$ p(x,y)=a x^3+b x^2y+c x y^2+d y^3 = 0 $$

making $\lambda = \frac xy$ and substituting we have

$$ (a\lambda^3+b\lambda^2+c\lambda + d)y^3 = 0 $$

so assuming $y \ne 0$ we have

$$ a\lambda^3+b\lambda^2+c\lambda + d = 0 $$

and for each pair $x_i,y_i$ we have

$$ a\lambda_i^3+b\lambda_i^2+c\lambda_i + d = 0 $$

now

$$ (1-\lambda_1)(1-\lambda_2)(1-\lambda_3) = 1 -(\lambda_1+\lambda_2+\lambda_3)+(\lambda_1\lambda_2+\lambda_2\lambda_3)-\lambda_1\lambda_2\lambda_3 $$

but by Vieta's formulas

$$ \frac ba = -(\lambda_1+\lambda_2+\lambda_3)\\ \frac ca = (\lambda_1\lambda_2+\lambda_2\lambda_3)\\ \frac da = -\lambda_1\lambda_2\lambda_3 $$

hence

$$ (1-\lambda_1)(1-\lambda_2)(1-\lambda_3) = 1+\frac ba+\frac ca+\frac da = 1+\frac{b+c+d}{a} $$

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  • $\begingroup$ I dont understand subt 4 part $\endgroup$ – user644938 Feb 16 at 4:02
  • $\begingroup$ @user644938 I hope now it is clear. $\endgroup$ – Cesareo Feb 16 at 14:01
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Hint: We substitute $$y=tx$$ then we get $$x^3(1-3t^2)=2010$$ $$x^3(t^3-3t)=2009$$ Dividing both equations we get $$2009(1-3t^2)=2010(t^3-3t)$$ This is a polynomial in degree three. Can you solve this equation?

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We have $$2009(x^3-3xy^2)=2010(y^3-3x^2y)$$ or $$2009x^3+6030x^2y-6027xy^2-2010y^3=0,$$ which by the Viete's theorem gives $$\left(1-\frac{x_1}{y_1}\right)\left(1-\frac{x_2}{y_2}\right)\left(1-\frac{x_3}{y_3}\right)=$$ $$=1+\frac{6030}{2009}-3-\frac{2010}{2009}=\frac{2}{2009}.$$

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