Find The Area Of The Equilateral Triangle

Show that the curve $x^3+3xy+y^3=1$ has only one set of three distinct points, $P$, $Q$, and $R$ which are the vertices of an equilateral triangle, and find its area.

My solution:

The first thing I notice is that there is cyclic symmetry between $x$ and $y$, and so setting $y=x$, we find:

[math]2x^3+3x^2-1=(x+1)^2(2x-1)=0[/math]

Thus, we know the points:

[math](x,y)=(-1,-1),\,\left(\frac{1}{2},\frac{1}{2} \right)[/math]

are on the given curve. Next, if we begin with the line:

[math]y=1-x[/math]

and cube both sides, we obtain:

[math]y^3=1-3x+3x^2-x^3[/math]

We may arrange this as:

[math]x^3+3x(1-x)+y^3=1[/math]

Since $y=1-x$, we may now write

[math]x^3+3xy+y^3=1[/math]

And since the point [math]\left(\frac{1}{2},\frac{1}{2} \right)[/math] is on the line $y=1-x$, we know the locus of the given curve is the line $y=1-x$ and the point $(-1,-1)$. Hence, there can only be one set of points on the given curve that are the vertices of any triangle, equilateral or otherwise.

Using the formula for the distance between a point and a line, we find the altitude of the equilateral triangle will be:

[math]h=\frac{|(-1)(-1)+1-(-1)|}{\sqrt{(-1)^2+1}}=\frac{3}{\sqrt{2}}[/math]

Using the Pythagorean theorem, we find that the side lengths of the triangle must be:

[math]s=\frac{2}{\sqrt{3}}h=\sqrt{6}[/math]

And so the area of the triangle is:

[math]A=\frac{1}{2}sh=\frac{1}{2}\sqrt{6}\frac{3}{\sqrt{2}}=\frac{3\sqrt{3}}{2}[/math]