I see no connection

First, we notice that $\dfrac{a^3+b^3+c^3}{3}-abc=\dfrac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{3}$

In order to minimize this expression, let's first see what we must do to minimize $a^2+b^2+c^2-ab-bc-ca$ (it's clear that to minimize $a+b+c$ we want $a,b,c$ as small as possible).

Now, WLOG let $a < b < c$ . This means $a=b-m$ and $c=b+n$ .

Plugging these substitutions in the second part that we wanted to minimize, we see that $a^2+b^2+c^2-ab-bc-ca=m^2+mn+n^2$ which is minimized when $m,n$ are minimized. Thus, $m=n=1$ .

Now we know that the three variables are in fact $b-1,b,b+1$ , so it remains to minimize $b-1+b+b+1=3b$ . But this is done by minimizing $b$ , which has a minimum of $b=2$ (since $b-1$ is a positive integer).

Thus, our expression attains minimum at $(a,b,c)=(1,2,3)$ which gives a minimum of $\dfrac{1^3+2^3+3^3}{3}-(1)(2)(3)=\boxed{6}$

I believe there was a typo, and it should say non-negative integers. In that case, we have $b=1$ which gives $(a,b,c)=(0,1,2)$ which gives a minimum of $\dfrac{0^3+1^3+2^3}{3}-(0)(1)(2)=\boxed{3}$

Note: The condition is pretty much useless, because the smallest possible values of $a,b,c$ gives $ab+bc+ca=(0)(1)+(1)(2)+(2)(0)=2\ge 2$

lol the answer is what you get when you pick the three smallest numbers possible. 1,2,and 3.