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Cubic conundrum (Posted on 2021-10-18) Difficulty: 2 of 5
Let x, y be real numbers other than 0 that satisfy x3+y3+3x2y2=x3y3. Find all possible values of 1/x+1/y.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

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Some Thoughts re: Solution | Comment 2 of 3 |
(In reply to Solution by Jer)

We can get that graph analytically.  

Start by substituting the identity x^3+y^3 = (x+y)^3-3xy(x+y) into the relation:
(x+y)^3-3xy(x+y)+3(xy)^2 = (xy)^3

Rearrange terms:
(x+y)^3 - (xy)^3 = 3xy(x+y) - 3(xy)^2

Factor:
(x+y-xy) * ((x+y)^2+(x+y)*(xy)+(xy)^2) = (x+y-xy) * (3xy)

Then either x+y-xy = 0 or (x+y)^2+(x+y)*(xy)+(xy)^2 = 3xy.
The first equation is the hyperbola 1/x+1/y=1.

The second equation seems to be a degenerate curve.  Trying to implicitly plot it yields two points (0,0) and (-1,-1).  Changing the 3 on the right side into something bigger like 3.1 forms a graph with an egg-shaped curve going around (-1,-1)

  Posted by Brian Smith on 2021-10-19 13:01:37
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