<s>[INST] Putnam 2006 B1 Problem 
Show that the curve $x^{3}+3xy+y^{3}=1$ contains only one set of three distinct points, $A,B,$ and $C,$ which are the vertices of an equilateral triangle, and find its area.

Yikes. Without knowing that this is the Folium of Descartes, it says the equation is reducible somehow...
$$x^3 + 3xy + y^3 - 1 = 0$$
Is factorable somehow. I tried the cubic way, but it is still insane... 
$$(x - 1)(x^2 + x + 1) + y(y^2 + 3x) = 0$$
But that doesnt help any bit, I could change $3xy$ I suppose:
$$\implies x^3 + xy + xy + xy + y^3 - 1 = 0$$
But that doesnt help either! 
 [/INST] Play around with it:  $$\begin{align*} 0 &= x^3 + 3xy + y^3 - 1 \\ &= (x+y)^3 - 3x^2 y - 3xy^2 + 3xy - 1 \\ &= (x+y)^3 - 1^3 - 3xy(x+y - 1) \\ &= (x+y-1)((x+y)^2 + (x+y) + 1) - 3xy(x+y-1) \\ &= (x+y-1)(x^2 - xy + y^2 + x + y + 1). \end{align*}$$
The first factor is the line $x + y  = 1$; the second is a conic section.  Under the rotation $$(x,y) = \left( \frac{u+v}{\sqrt{2}}, \frac{u-v}{\sqrt{2}} \right),$$ we obtain for this factor $$\begin{align*}  0 &= x^2 - xy + y^2 + x + y + 1 \\ &= (u + \sqrt{2})^2 + 3v^2. \end{align*}$$  As $u, v \in \mathbb R$, it follows that the only solution is $(u,v) = (-\sqrt{2}, 0)$, corresponding to $(x,y) = (-1,-1)$; thus this conic is degenerate.  Since it is not possible for three collinear points to form an equilateral triangle, one vertex of any such triangle must be at this isolated point.  The other two points must lie on the aforementioned line; it is now straightforward to show that the other two points must be unique, being the intersection of a circle with some radius $r$ centered at $(-1,-1)$ and the line.  The rest I leave as a simple exercise.
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