Show that the equation of the folium of Descartes in terms of $x$ and $y$ is $x^3+y^3=axy$

Question

I'm given that the parametric equations are

$x=\frac{at}{1+t^3}$ and $y=\frac{at^2}{1+t^3}$ and that $a>0$

Here's my attempt at a solution:

Find $x^3$ and $y^3$ in terms of $t$..

$x^3=\left(\frac{at}{1+t^3}\right)^3$ and $y^3=\left(\frac{at^2}{1+t^3}\right)^3$

Add them together:

$x^3+y^3=\left(\frac{at}{1+t^3}\right)^3+\left(\frac{at^2}{1+t^3}\right)^3$

Since we're trying to prove that the above is equal to $axy$, we find $axy$ in terms of $t$ which is:

$axy=a\left(\frac{at}{1+t^3}\right)\left(\frac{at^2}{1+t^3}\right)$

Then finally to prove $x^3+y^3=axy$ I have to prove:

$a\left(\frac{at}{1+t^3}\right)\left(\frac{at^2}{1+t^3}\right) = \left(\frac{at}{1+t^3}\right)^3+\left(\frac{at^2}{1+t^3}\right)^3$

And that's where I'm stuck. I can't figure out why those two things are equal.

Answer 1

In $$a\left(\frac{at}{1+t^3}\right)\left(\frac{at^2}{1+t^3}\right) = \left(\frac{at}{1+t^3}\right)^3+\left(\frac{at^2}{1+t^3}\right)^3,$$ the left side is $a\left(\dfrac{at}{1+t^3}\right)\left(\dfrac{at^2}{1+t^3}\right) =\dfrac{a^3t^3}{(1+t^3)^2} $ and the right side is

$\begin{align} \left(\frac{at}{1+t^3}\right)^3+\left(\frac{at^2}{1+t^3}\right)^3 &=\frac{a^3t^3}{(1+t^3)^3}+\frac{a^3t^6}{(1+t^3)^3}\\ &=\frac{a^3t^3+a^3t^6}{(1+t^3)^3}\\ &=\frac{a^3t^3(1+t^3)}{(1+t^3)^3}\\ &=\frac{a^3t^3}{(1+t^3)^2}\\ \end{align} $

so the two are equal.

Answer 2

Find t first by dividing the 1st equation by the second equation (note that $1+t^3$ will cancel each other):

$x/y = at/at^2$

$x/y = 1/t$

$t = y/x.$

Now, substitute this to the 1st equation, simplify and you will be done :)