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 Given Cubic, Solve For Real (Posted on 2024-01-10)
Solve this equation in real numbers:

(x+y)3 = z
(y+z)3 = x
(z+x)3 = y

 No Solution Yet Submitted by K Sengupta No Rating

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 Special case Comment 1 of 1
If x=y=z = t then (2t)^3 = t
t(8*t^2 - 1) = 0
t = {0, √2/4, -√2/4}

Three solutions are:
(x,y,z) = {(0,0,0), (√2/4,√2/4,√2/4), (-√2/4,-√2/4,-√2/4)}

There may be other solutions, but I consider it feasible that these are the only solutions.

Reasoning:  Looking at any one of the equations, in 3D it looks similar to the 2 dimensional equation y=x^3 except the pattern is extended above and below "the paper", as if you took a flat plane then bent it into a y=x^3 shape when viewed on end.  Only rotated relative to the x,y,z axes.
Now intersect 2 orthogonal versions of that structure and you would get a line curving through 3 dimensions.
Intersect that with one more orthogonal structure and I think you would get a finite number of points.

Obviously there may be more than just these 3 points, I look forward to more solutions or a proof that these are the full solution set.

 Posted by Larry on 2024-01-10 11:18:15

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