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Question Number 68063 by TawaTawa last updated on 04/Sep/19

Answered by MJS last updated on 04/Sep/19

$$\mathrm{tricky}\mathrm{but}\mathrm{easy}$$$$\left(1\right)x^3-3y^{2}x=a+1$$$$\left(2\right)y^3-3x^{2}y=a$$$$\mathrm{let}1-\frac{x}{y}=z\iff y=\frac{x}{1-z}\mathrm{this}\mathrm{leads}\mathrm{to}$$$$\left(1\right)x^3\frac{z^2-2z-2}{{(z-1)}^2}=a+1\Rightarrow x^3=\frac{(a+1){(z-1)}^2}{z^2-2z-2}$$$$\left(2\right)x^3\frac{3z^2-6z+2}{{(z-1)}^3}=a\Rightarrow x^3=\frac{a{(z-1)}^3}{3z^2-6z+2}$$$$\Rightarrow$$$$\frac{(a+1){(z-1)}^2}{z^2-2z-2}=\frac{a{(z-1)}^3}{3z^2-6z+2}$$$$\mathrm{transforming}\mathrm{to}$$$$z^3-\frac{3(2a+1)}{a}{z}^2+\frac{6(a+1)}{a}z-\frac{2}{a}=0$$$$\mathrm{let}\alpha ,\beta ,\gamma \mathrm{are}\mathrm{the}\mathrm{solutions}\mathrm{of}\mathrm{this}$$$$\Rightarrow$$$$(z-\alpha )(z-\beta )(z-\gamma )=0$$$$z^3-(\alpha +\beta +\gamma )z^2+(\alpha \beta +\alpha \gamma +\beta \gamma )z-\alpha \beta \gamma =0$$$$\Rightarrow \mathrm{by}\mathrm{comparing}\mathrm{factors}\alpha \beta \gamma =\frac{2}{a}$$$$\mathrm{but}\alpha \beta \gamma =z_1{z}_2{z}_3=(1-\frac{x_1}{y_1})(1-\frac{x_2}{y_2})(1-\frac{x_3}{y_3})=\frac{2}{a}$$$$\mathrm{in}\mathrm{our}\mathrm{case}a=2009$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}\frac{2}{2009}$$

Commented by TawaTawa last updated on 04/Sep/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Learner-123 last updated on 04/Sep/19

$$Amazing!$$

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