Locus of all point $P(x,y)$ satisfying $x^3+y^3+3xy=1$

Question

Locus of all point $P(x,y)$ satisfying $x^3+y^3+3xy=1$ consists of union of
$(A)$a line and an isolated point
$(B)$a line pair and an isolated point
$(C)$a line and a circle
$(D)$a circle and an isolated point.

How do we factorize this to decide whose combined equation is this?Or is there some other method by which we can decide without factorizing?I could not factorize the equation.Any suggestions?

Answer 1

Notice this cool factorization $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=\frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(a-c)^2)$$ Apply it for $a=x, b=y, c=-1$ to obtain $$x^3+y^3+3xy-1=\frac{1}{2}(x+y-1)((x-y)^2+(x+1)^2+(y+1)^2)$$ and you surely know how to conclude from here!

Answer 2

$$\begin{align}x^3+y^3+3xy-1&=(x+y)^3-3xy(x+y)+3xy-1\\&=(x+y)^3-1-3xy(x+y-1)\\&=(x+y-1)((x+y)^2+(x+y)+1)-3xy(x+y-1)\\&=(x+y-1)((x+y)^2+x+y+1-3xy)\\&=(x+y-1)(x^2+y^2-xy+x+y+1)\end{align}$$ Note here that $$\begin{align}x^2+y^2-xy+x+y+1&=x^2+(1-y)x+y^2+y+1\\&=\left(x+\frac{1-y}{2}\right)^2-\left(\frac{1-y}{2}\right)^2+y^2+y+1\\&=\left(x+\frac{1-y}{2}\right)^2+\frac 34(y+1)^2\end{align}$$