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

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SUMMARY

The curve defined by the equation x^3 + 3xy + y^3 = 1 contains exactly one set of three distinct points, A, B, and C, which form the vertices of an equilateral triangle. Through analysis, it was determined that all points except (-1, -1) lie on the line y = 1 - x. The equation can be factored into a useful form, revealing the structure of the curve and confirming the unique solution set. The points (-1, -1), (-1, 2), and (2, -1) were identified as the vertices of the equilateral triangle.

PREREQUISITES
  • Understanding of polynomial equations and their factorizations
  • Familiarity with the properties of equilateral triangles
  • Basic knowledge of coordinate geometry
  • Experience with plotting points on a Cartesian plane
NEXT STEPS
  • Study polynomial factorization techniques in depth
  • Explore the geometric properties of equilateral triangles
  • Learn about implicit functions and their graphical representations
  • Investigate advanced methods for solving cubic equations
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Mathematics students, educators, and anyone interested in algebraic geometry or the properties of geometric shapes will benefit from this discussion.

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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).
 

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