WLOG $a\geq b \geq c$ (symmetric equation). It is easy to see that $a \geq 9$ while $c \leq 8$. we also know that $x^3 \equiv 0,1,8 \pmod 9$ and as $a^3 + b^3 + c^3 \equiv 0 \pmod 9$, we get that either they all are divisible by $3$, or one of them is and the other ones are $\equiv 1,8 \pmod 9$. Notice that if they all are divisible by $3$, then $27|a^3 + b^3 + c^3$ but $27\not \mid 2016$ so not possible. Therefore only one of them is divisible by $3$. We also notice that $13^3 > 2016$ so $a \leq 12$. Therefore, $a=9,10,11,12$. If $a=9$, then $b =9$ because if $b\leq 8$, then $729 + 512 + c^3 \geq a^3 + b^3 + c^3 = 2016 \iff c^3 \geq 800$ approx which isn't possible as $c\leq b$. So $a=9$ implies $b=9$ but we already discussed that this case isn't possible. Now $a = 12$ gives $2016 - 1728 = 288 = b^3 + c^3$ which implies $6 \leq b < 7$ but $a=12,b=6$ isn't possible as they both are divisible by $3$. $a=11$ gives $685 = b^3 + c^3$ implying $7 \leq b \leq 8$. This also means $c=3,6$ as atleast one of them has to be divisible by $3$ and clearly this doesn't give any solution. Now $a=10$ gives $1016 = b^3 + c^3$ meaning $9 \leq b \leq 10$. $b=10$ clearly not possible so $b=9$ meaning $c=287$ which isn't possible. Therefore no solution.

Taking mod $7$, we have $a^3+b^3+c^3\equiv 0\pmod{7}$. Now, examine the third power. All the possible third power modulo $7$ are $0, 1, 1, -1, 1, -1, -1$. Thus, the only way is to have $1, -1, 0$ as a combination. With this constraint, plug in one of them to be the exponent of $0\pmod{7}$, which can be either $0$ or $7$. Now, split into casework gives $b^3+c^3=2016$ or $b^3+c^3=1673$. Now if $b^3+c^3=2016$, then we have no solutions by bounding since $b^3+c^3=(b+c)((b+c)^2-3bc)$ which we can let $b+c=m, bc=n$. This gives $(m)(m^2-3n)=2016$ and both $m, n$ are integers. Try factors of $m$ yields no solutions. Furthermore, if $b^3+c^3=(b+c)(b^2+c^2-bc)=1673$, then since both are integers one of them must be $7$ and the other be $239$, which is a prime. This clearly yields no solutions. Thus, there are no such triple. q.e.d.

Solution:

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In the above solution, Would anyone enlighten me how to show $3a^3 \geq a^3 +b^3 +c^3?$

Vulch wrote: In the above solution, Would anyone enlighten me how to show $3a^3 \geq a^3 +b^3 +c^3?$ Because we wlog $a\geq b\geq c\implies a^3\geq b^3\geq c^3$.

Vulch wrote: In the above solution, Would anyone enlighten me how to show $3a^3 \geq a^3 +b^3 +c^3?$ Bounding.