Eliminate the parameter from the parametric equations $$x=\frac{3t}{1+t^3} ,  y=\frac{3t^2}{1+t^3}  , t \neq -1,$$
and hence find an ordinary equation in x and y for this curve, The parameter t can be interpreted as the slope of the line joining the general point $(x,y)$ to the origin. Sketch the curve and show that the line $x+y=-1$ is an asymptote.

$$ x^3 + y^3 =\frac{27t^3}{(1+t^3)^3} +\frac{27t^6}{(1+t^3)^3}
=\frac{27t^3(1+t^3)}{(1+t^3)^3}=\frac{27t^3}{(1+t^3)^2}$$
$$ x^3 + y^3 = 3xy$$
