On the Diophantine equation $2(a^3+b^3+c^3)=abc(abc+6)$ 
I am sure that $(3, 2, 1)$ is the only natural triple satisfying the following Diophantine equation $$2(a^3+b^3+c^3)=abc(abc+6)\quad\text{for}\quad a\ge b\ge c.$$
  But I can not prove or refute that. Help me, please.

What I've tried so far. Rearrange the equation to get
$$
2(a^3+b^3+c^3-3abc)=a^2 b^2 c^2
$$
and
$$
(a+b+c)((a-b)^2+(a-c)^2+(b-c)^2)=a^2 b^2 c^2
$$
Let $a-b=n$ and $a-c=m$. Hence $m \ge n$ and
$$
(3a-m-n)(m^2+n^2+(m-n)^2)=[a(a-m)(a-n)]^2
$$
or
$$
2(3a-(m+n))((m+n)^2-3mn)=[a(a^2-(m+n)a+mn)]^2
$$
Now let $m+n=p$ and $mn=q$, then we have
$$
2(3a-p)(p^2-3q)=[a(a^2-pa+q)]^2
$$
Now we have a quadratic in $q$ and I tried to use $\Delta_{q}=k^2$ (where $k$ is an integer) for bounding some of variables, but I was unsuccessful. 
 A: HINT
Using the identity
$$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$
one can get the original equation in the form of
$$2(a+b+c)((a+b+c)^2-3(ab+bc+ac)) = (abc)^2,$$
or
$$r=\frac{2s^3-p^2}{6s},$$
where
$$s=a+b+c,\quad r=ab+bc+ac,\quad p=abc,$$
$$a>0,\quad b>0,\quad c>0,\quad s^3-27p\ge 0.$$
That allows to write the cubic equation for $x\in\{a, b, c\}$ in the form of
$$x^3 - sx^2 + \frac{2s^3-p^2}{6s}x - p = 0,$$
or
$$6sx^3 - 6s^2x^2 + (2s^3-p^2)x - 6sp =0.$$
For natural solutions
$$p=qx,$$
$$(6s-q^2)x^2 - 6s^2x + 2s(s^2-3q) = 0.\qquad(1)$$
The "linear" case in $x$ is
$$\begin{cases}
q^2 = 6s\\
x_1=\dfrac{s^2-3q}{3s}\\
x_{2,3}^2 - (s-x_1)x_{2,3} + q = 0\\
x>0,\quad s>0,\quad q>0.
\end{cases}\qquad(2)$$
System $(2)$ gives the only solution:
$$x_1=\frac{q^2}{18} - \frac 6q,\quad q=s=6,\quad x\in\{1,2,3\},$$
$$\boxed{(a,b,c) = (3,2,1)}.$$
The "quadratic" case in $x$ gives:
$$\begin{cases}
6s\not= q^2\\
x_1 = \dfrac{3s^2\pm d}{6s-q^2}\\
d^2 = 9s^4 - 2s(s^2-3q)(6s-q^2)\\
x_{2,3}^2 - (s-x_1)x_{2,3} + q = 0\\
x>0,\quad q>0,\quad s^3\ge 27qx_1.
\end{cases}\qquad(3)$$
UPD
Using AM-GM inequality:
$$x_1 = s-(x_2+x_3)\ge s-2\sqrt{x_2x_3},$$
$$x_1\ge s-2\sqrt q,$$
so
$$\dfrac{2s(s^2-3q)}{3s^2\mp d}\in\mathcal N,\quad \dfrac{2s(s^2-3q)}{3s^2\mp d}\ge s-2\sqrt q\ge0,\quad s^2\ge4q.\qquad(4).$$
Attempts to obtain other solutions either of $(1-4)$ in the case of $6s\not= q^2\\$ do not lead to a positive result. One can do substution
$$s=w\sqrt q,$$
obtaining
$$6w\not=q\sqrt q,\quad\dfrac{2w(w^2-3)}{3w^2\mp\sqrt{9w^4 - 2w(w^2-3)(6w-q\sqrt q)}}\ge w-2\ge0$$
and closed inequality in form $q(w)\ge0$, but even in the case 
$$6w<q\sqrt q,\quad \text{"}\mp\text{"}=\text{"}+\text{"}$$
one get infinity set of inequality solutions, so this don't change the situation cardinally.
This gives grounds for considering that solution $(3,2,1)$ is an easy case of the OP task and reduces the chances of success in the general case.
