Let $ P(a,b,c) = a^2b+b^2c+c^2a-ab^2-bc^2-ca^2 $. Notice that if $ a=b $, $ P(a,b,c) = a^3+a^2c+ac^2-a^3-ac^2-a^2c = 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\cdot5\cdot4 = 120 $ triples where $ a,b,c $ are all distinct, and $ 6\cdot6\cdot6=216 $ triples total, so there are $ 216-120 = \boxed{96} $ triples such that $ P(a,b,c) = 0 $.