<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] Hint:
$$0=a^3-b^3-c^3-3abc = \frac12(a-b-c)[(a+b)^2+(b-c)^2+(c+a)^2]$$
So we have two cases:
Case 1:
$a=b+c \implies a^2=2a \implies a \in \{0, 2\}$.
You should be able to fnd out $b+c$ bounded in this case...
OR Case 2:
$a+b=b-c=c+a=0$, which again reduces cases considerably among natural numbers.
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