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Taking the Fifth (Posted on 2014-06-21) Difficulty: 3 of 5

For the most part, if z^5 is the sum of two positive cubes, then z^2 is itself the sum of positive cubes:
(z^2 = x^3 + y^3, so z^2*z^3 = z^5 = (z*x)^3 + (z*y)^3).

Hence, such solutions to z^5=x^3+y^3 are said to be 'trivial'

In this sense, 3549^5 is the non-trivial sum of two positive cubes. It is nevertheless possible to calculate the cubes without resort to brute force.

How, and what are they?

No Solution Yet Submitted by broll    
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  Subject Author Date
re(2): SolutionBrian Smith2017-05-05 23:54:15
re: Solutionbroll2017-05-05 09:20:26
SolutionSolutionBrian Smith2017-05-01 23:55:28
Hints/TipsAnother cluebroll2014-07-04 08:46:47
Hints/TipsHintbroll2014-06-22 10:24:33
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