jmc

Let $P(a,b,c)=a^{2}b+b^{2}c+c^{2}a-a{b}^2-b{c}^2-c{a}^2$. Notice that if $a=b$, $P(a,b,c)=a^3+a^{2}c+a{c}^2-a^3-a{c}^2-a^{2}c=0$. By symmetry, $P(a,b,c)=0$ when $b=c,c=a$ as well. Since $P(a,b,c)$ has degree 3 and is divisible by three linear terms, $P(a,b,c)$ must factor as $k(a-b)(b-c)(c-a)$ where $k$ is a constant. Hence, $P(a,b,c)=0$ if and only if at least two of $a,b,c$ are equal.

To count how many triples $(a,b,c)$ satisfy this, we count the complement. There are $6\cdot 5\cdot 4=120$ triples where $a,b,c$ are all distinct, and $6\cdot 6\cdot 6=216$ triples total, so there are $216-120=\boxed{96}$ triples such that $P(a,b,c)=0$.