Algebra problem: a^3 + b^3 + c^3 + (a+b+c)^3 = 2016

[John L]

Hello , can someone help me ?


a^3 + b^3 + c^3 + (a+b+c)^3 = 2016


a,b,c- positive integers

(a,b,c)=?

I got stuck there :

2(a^3 + b^3 + c^3) + 3(a+b)(b+c)(a+c) = 2016

 

[John L]

This problem is from a math gazette called RMT

 

[John L]

You sure seem desperate to get solution:
cheating on a competition problem?

http://www.sosmath.com/CBB/viewtopic.php?f=1&t=70432

Nah i just want to know if you can solve this problem without saying that 13^2 > 2017 and then guesing the numbers

Oh and that gazette is called "revista matematica timisoara" and you can find it only in my country

 

[Deleted member 4993]

Nah i just want to know if you can solve this problem without saying that

13^2 > 2017 → Incorrect..... 13^2 = 169 which is surely less than 2017

and then guesing the numbers

Oh and that gazette is called "revista matematica timisoara" and you can find it only in my country

.

 

[John L]

.

13^3 sorry