What is the solution to a^3+b^3+c^3 = 495, a+b+c = 15, and abc = 105?

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Discussion Overview

The discussion revolves around solving the system of equations involving the cubes of three variables: \(a^3 + b^3 + c^3 = 495\), \(a + b + c = 15\), and \(abc = 105\). The focus is on exploring potential solutions, particularly under the assumption of integer values for the variables.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests starting by cubing the second equation and substituting the first and second equations into the new expression to find a polynomial in \(a\).
  • Another participant proposes that assuming whole number solutions simplifies the problem, noting that the product \(abc = 105\) limits the possible integer combinations.
  • A specific combination of integers (3, 5, 7) is identified as a potential solution, as they sum to 15 and their cubes add up to 495.
  • A later reply supports the assumption of integer values, arguing that it significantly reduces the complexity of finding solutions.

Areas of Agreement / Disagreement

Participants generally agree on the approach of seeking integer solutions, but the discussion does not resolve whether the identified integers are the only solutions or if other combinations exist.

Contextual Notes

The discussion assumes integer values for \(a\), \(b\), and \(c\), which may limit the exploration of other potential solutions. The method of cubing the second equation and substituting may involve unresolved mathematical steps.

Who May Find This Useful

Readers interested in algebraic problem-solving, particularly those focused on systems of equations and integer solutions, may find this discussion relevant.

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solve :
a^3+b^3+c^3 = 495
a+b+c = 15
abc = 105
thanx
regards
abc
 
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I would probably start by cubing the second equation, and then I'd substitute the first and second equation wherever possible in the new expression. After that I'd look for a possible factorisation of the equation so that I can substiute b + c and bc in terms of a. Hopefully this will yield a simple (that is, a low degree) polynomial in a. And similarly for the other variables (after the value of a has been determined).
 
Assume a whole number solution:
abc=105 means then that only 1,3,5,7,105 can be solutions;
and only 3,5,7 can be summed to 15
7^3=343
5^3=125
3^3=27
Sum=495
 
Arildno's assumption is acceptable (and in this case results in saving a lot of time)because the odds of finding integer values for the three expressions get vanishingly small unless a,b,c are integers.

And it's easy to narrow down to the choice of 3,5,7 for getting sum = 15, by noticing how close 15 is to thrice the cube root of 105. You probably get it faster by inspection.
 
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