I am asked to find Area of the equilateral triangle formed by $(x,y)$ satisfying given curve $x^3+y^3+3xy=1$
I can find no way to find the points on this curve that satisfy an equilateral triangle. Looking at the polynomial I get $(0,1)$, $(1,0)$ and $(-1,-1)$ as its solutions, don't know how that could possibly help.
I can't even find a way to reduce this polynomial using the traditional formulae
Try this. $$x^3 + y^3 + z^3 - 3xyz = \dfrac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]$$
So, for our problem \begin{equation} x^3+y^3 + 3xy = 1\\ \Rightarrow\quad \dfrac{1}{2}(x+y-1)[(x-y)^2+(y+1)^2+(x+1)^2] = 0\\ \end{equation} Since $\dfrac{1}{2}[(x-y)^2+(y-z)^2+(z-x)^2] > 0$, if $x = y \neq -1$, this is an equation for the straight line $x+y+1=0$ and a single point $(-1,-1)$.
We can find the area from here.