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The question

If the tangent at $(x_1,y_1)$ to the curve $x^3+y^3=a^3$ meets the curve again at a point $(x_2,y_2)$ then prove that $$\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$$

My attempt

By differentiating the above expression and finding the slope, the equation of the tangent through the point $(x_1,y_1)$ is $$yy_1^2+xx_1^2=y_1^3+x_1^3$$

As it passes through the point $(x_2,y_2)$, $$y_2y_1^2+x_2x_1^2=y_1^3+x_1^3$$

But I am unable to get the required condition from here.

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Given $x^3+y^3=a^3$

Then find $\frac{dy}{dx}=-\frac{x^2}{y^2}$

Then the slope at $(x_1.y_1)$ is $$-\frac{x_1^2}{y_1^2}$$

The slope of the tangent passing through $(x_2,y_2)$ is $$\frac{y_2-y_1}{x_2-x_1}$$

Now lets compare the two equations and we get $$\frac{y_2-y_1}{x_2-x_1}=-\frac{x_1^2}{y_1^2}$$ $$\frac{y_2^3-y_1^3}{x_2^3-x_1^3}\left(\frac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}\right)=-\frac{x_1^2}{y_1^2}$$ $$-\frac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}=-\frac{x_1^2}{y_1^2}$$ $$x_1^2y_1^2+x_1x_2y_1^2+x_2^2y_1^2=x_1^2y_1^2+x_1^2y_1y_2+x_1^2y_2^2$$ $$x_1^2y_2^2-x_2^2y_1^2=x_1x_2y_1^2-x_1^2y_1y_2$$ $$(x_1y_2-x_2y_1)(x_1y_2+x_2y_1)=x_1y_1(x_2y_1-x_1y_2)$$ $$x_1y_2+x_2y_1=-x_1y_1$$ $$\frac{x_2}{x_1}+\frac{y_2}{y_1}=-1$$ $$\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$$

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