Proving Equilateral Triangle from x^3+3xy+y^3=1

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The equation x^3 + 3xy + y^3 = 1 describes a curve that contains three distinct points A, B, and C, which are the vertices of an equilateral triangle. Initial attempts to find points on the curve revealed that most lie on the line y = 1 - x, except for the point (-1, -1). The equation can be factored into a useful form that simplifies the analysis of the curve's properties. By examining the roots, the points (-1, -1), (-1, 2), and (2, -1) were identified, although the point (-1, -1) presents issues due to non-derivability. Ultimately, the discussion emphasizes the need for a strategic approach to factorization when tackling such problems.
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Homework Statement


Show that the curve x^3+3xy+y^3=1 contains only one set of three distinct points A,B, and C, which are vertices of an equilateral triangle.

Homework Equations





The Attempt at a Solution


I randomly starting plotting points and found that all of them fell on the line y=1-x except (-1,-1). So, I just need to prove that these are the only points that satisfy that equation. It turns out that that equation factors into a very useful form. See B1 http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2006s.pdf
My question is how would you "discover" that really nice form if you were taking the test?
 
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The first insticnt would be to try to factorise.

x^3 + y^3 +3xy -1
= (x+y)^3 -3x^2y -3xy^2 +3xy -1
= [(x+y)^3 -1] -3xy(x+y-1)
= [(x+y-1){(x+y)^2 +(x+y) +1}] -3xy(x+y-1)

Now you can take (x+y-1) common and the rest would follow.
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p.s.

If a simpler method occurs to me, I'll immdtly let you know. By inspection, some roots can be found to be (-1,-1), (-1,2) and (2,-1). It's not derivable at (-1,-1). If we plot the three points, we can sense some trouble at (-1,-1).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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