Bro , do you even Euler?

The second equation is equivalent to $a^3 = 2a(b+c)$ . We can substitute that into the first equation to get $2a(b+c) - b^3 - c^3 = 3abc$ $2a(b+c) - 3abc = b^3 + c^3$ $a = \frac{b^3 + c^3}{2b+2c - 3bc}$ Since $a > 0$ and $b^3 + c^3 > 0$ , we must also have $2b+2c-3bc > 0$ . Rewriting this inequality a little gives us: $bc - \frac{2}{3}(b+c) + \frac{4}{9} < \frac{4}{9}$ $\left(b - \frac{2}{3}\right)\left(c - \frac{2}{3}\right) < \frac{4}{9}$ If $b = c = 1$ then this condition is met. However, if at least one of them is at least $2$ , then we have $\left(b - \frac{2}{3}\right)\left(c - \frac{2}{3}\right) \ge \left(2 - \frac{2}{3}\right)\left(1 - \frac{2}{3}\right) = \frac{4}{9}$ . This is a contradiction since it implies $\frac{4}{9} > \frac{4}{9}$ .

Hence the only possible solution occurs when $b = c = 1$ . Then $a = 2$ , so the answer is $\boxed{4}$ .

A nice way to do this problem is to factor like this: $\displaystyle { a }^{ 3 }-{ b }^{ 3 }-{ c }^{ 3 }-3abc=(a-b-c)({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+ab+ac-bc)$ As ${ a }^{ 3 }-{ b }^{ 3 }-{ c }^{ 3 }-3abc=0$ , we have that $a=b+c$ or another expression that makes the remaining factor zero. Luckily, the remaining factor can be expressed as (with quite a bit of work lol) $\displaystyle { \left( \frac { a-b }{ 2 } +c \right) }^{ 2 }+\frac { 3 }{ 4 } { (a+b) }^{ 2 }$ Which means you can just ignore it, since it can only be zero if you use 0. This is because with integers $a+b$ must be $0$ since both are squares, so one of $a,b$ is not a natural number. Interestingly, this makes the other square $a+c$ which implies $\left| a \right| =\left| b \right| =\left| c \right|$ if we're using integers to solve this particular expression! (More specifically in the form (a,-a,-a))

The system is equivalent to (where $k=b+c$ ) $\displaystyle \begin{cases} a=k \\ { a }^{ 2 }=2k \end{cases}$ So $k=2$ , and $a=2$ . Thus, $a+b+c=4$ since there is only one way for natural numbers to sum to two. ( $(a,b,c)=(2,1,1)$ )

A little different approach!!

First expression implies that $a-b-c=0$ giving $a=b+c$ .Substituting this in second equation gives values of $a$ as $0$ or $2$ .Therefore $a=2$ .Again we know that $a=2=b+c$ giving $(b,c)=(1,1)$ as the only pair.Therefore only triplet satisfying this system is $(2,1,1)$ .

ovberrated..so simple from 1st

a=b+c

sub into 2nd

aa=2a a=2 from conditions.

only possibiity

(2,1,1)