Putnam 2006 B1

$x^3+3xy+y^3-1=0$ The given equation resembles the algebraic identity , $\begin{aligned} a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\ &=(a+b+c)\cdot \frac{1}{2}\left [(a-b)^2+(b-c)^2+(c-a)^2\right ] \end{aligned}$ Substituting $a=x,\; b=y$ and $c=-1$ , we have, $\begin{aligned} x^3+y^3+(-1)^3-3xy(-1)&=0 \\ (x+y-1)\cdot \frac{1}{2}\left[(x-y)^2+(x+1)^2+(y+1)^2\right]&=0 \end{aligned}$ Therefore, the curve represents all points $(x,y)$ satisfying $x+y=1$ or $(x-y)^2+(x+1)^2+(y+1)^2=0\implies (x,y)=(-1,-1)$ .

To form a triangle, one of the vertices must be $A(-1,-1)$ and the other two vertices lie on the line $x+y=1$ .

The height of the equilateral triangle is $h=\sqrt{2}+\dfrac{\sqrt{2}}{2}=\dfrac{3\sqrt{2}}{2}$

The area of $\triangle ABC$ is $\dfrac{ h^2\sqrt{3}}{3}=\boxed{\dfrac{3\sqrt{3}}{2}}$