Root Bombs!

Let $a+b+c=ab+bc+ca=3abc=p$ . Then $a, b, c$ are the roots of the equation $x^3-px^2+px-\dfrac{p}{3}=0$ . This equation will have real roots if $-27\times {\dfrac{p^2}{9}}+18\times (-p) \times p\times (-\dfrac{p}{3})-4p^3-4\times (-p^3)\times (-\dfrac{p}{3})+p^2\times p^2\geq 0$ . This yields $(p-3)^2\leq 0$ . Since $(p-3)^2$ can't be negative, therefore $p-3=0$ or $p=3$ . So $3abc=3$ or $abc=\boxed 1$

Without loss of generality, let $a \leq b \leq c$ .

If $a=0$ , we have $b+c=bc=0$ ; so $b=c=0$ as well.

If $a=1$ , we have $1+b+c=b+c+bc$ , so that $bc=1$ and $b=c=1$ ; checking the remaining equation, we see that this is indeed a solution.

Now assume $a>1$ . Note that $ab \leq ca \leq bc$ , so $ab+bc+ca \leq 3bc < 3abc$ ; hence no other solutions exist.

Hence the total of all valid $abc$ is $\boxed1$ .

We can write 3abc(a+b+c)=(ab+bc+ca)². Now let bc=x,ca=y,ab=z, our equation transforms to (x+y+z)²=3(xy+yz+zx)=>x²+y²+z²-xy-yz-zx=0. Clearly, x=y=z=>a=b=c.

Let us start by solving for $c$ in terms of $a$ and $b$ :

$\frac{a+b}{3ab-1} = c = \frac{ab}{3ab - (a+b)} \Rightarrow (3b - 3b^2 - 1)a^2 + (3b^2 - b)a - b^2 = 0$ .

Solving for $a$ via the Quadratic Formula gives:

$a = \frac{(b-3b^2) \pm \sqrt{(3b^2 - b)^2 - 4(3b-3b^2-1)(-b^2)}}{2(3b - 3b^2 -1)} = \frac{(b-3b^2) \pm \sqrt{-3b^4 + 6b^3 - 3b^2}}{2(3b - 3b^2 -1)} = \frac{(b-3b^2) \pm \sqrt{-3b^2(b-1)^2}}{2(3b - 3b^2 -1)}$ (i).

The only way (i) will be real-valued is for $b = 0, 1$ , which in turn yield $a = 0, 1$ and $c = 0, 1$ respectively. These ultimately produce the two products:

$abc = 0^3$ or $abc = 1^3 \Rightarrow 0 + 1 = \boxed{1}.$