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

[abc]

solve :
a^3+b^3+c^3 = 495
a+b+c = 15
abc = 105
thanx
regards
abc

 

[Muzza]

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).

 

[arildno]

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

 

[Gokul43201]

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.

 

[s0l0m0nsh0rt]

OMG...LOL!


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