<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] use
\begin{align*}
a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\\
&=\dfrac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]
\end{align*}
if $a^3+b^3+c^3-3abc=0$ so we have $a=b=c$ or $a+b+c=0$
$$x^3+y^3+3xy=1\Longrightarrow x^3+y^3+(-1)^3-3xy\cdot (-1)=0$$
</s>