<s>[INST] Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers. 
Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers.

Substituting $a=\sqrt{2(b+c)}$ in the cubic equation, we get:
$2\sqrt{2}(b+c)^{\frac{3}{2}} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$
$2\sqrt{2}(b+c)\sqrt{b+c} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$
Not able to proceed further, although I think that binomial expansion of $\sqrt{b+c}$ can yield some more steps. But, am confused about applying that too.
 [/INST] \begin{align}x^3+y^3+z^3-3xyz & = x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2 \\
& =(x+y)^3+z^3-3xy(x+y+z)\\
&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z) \\ & =(x+y+z)(x^2+2xy+y^2+z^2-xy-xz-3xy) \\ & =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\end{align}
Therefore, for distinct $x,y $ and $z$; $$
x^3+y^3+x^3=3xyz \implies x+y+z=0$$ Notice that your equation can rewritten as  $$(a)^3+(-b)^3+(-c)^3=3 (a) (-b)(-c)$$
$$\implies a-b-c=0$$
Substitute $a=b+c$ in your second equation,
$$ 
a^2=2(b+c) \implies (b+c)^2=2(b+c) \implies b+c=2$$
So finally , we've $a=2$ and $b+c=2$.
So the only natural solution to these equations is  $\color{blue}{(a,b,c)=(2,1,1)}$.
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