Find locus of the points $(x,y)$ such that $x^3+y^3+3xy=1$

Question

Find locus of the points $(x,y)$ such that $x^3+y^3+3xy=1$

$\textbf{My solution}:$ We notice that $(x+y)^3=x^3+y^3+3xy(x+y)$, putting $x+y=1$ gives us our desired equation. Hence all the points satisfying $x+y=1$ is gives us all the points satisfying our original equation, which results in a linear graph. My concern about my solution is that the actual solution given in the book looks way more different and complicated than mine, so I doubt there is fallacy in my solution. Someone please point out if there's any.

You have found a solution. Now show that it is the only solution. — bobeyt6, Jul 11 '23 at 21:11
What you proved was that $x+y=1\implies x^3+y^3+3xy=1$. Is that what is proven in your book? Or is it the stronger statement $x+y=1\iff x^3+y^3+3xy=1$? — José Carlos Santos, Jul 11 '23 at 21:16

score 2 · Answer 1 · answered Jul 11 '23 at 21:34

Your solution misses the point $(-1,-1)$, but that’s it. Your first step in writing in terms of $(x+y)$ was excellent. You correctly saw that if $x+y=1$ then the equation is satisfied. You have found a linear factor of your cubic, namely $1-(x+y)$ so you can write it as that factor times a quadratic.

The full solution then is the zeros of that quadratic union the zeros of that linear term.

Find locus of the points $(x,y)$ such that $x^3+y^3+3xy=1$

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