Let \(0 \le a, b, c \le 5\) be integers. For how many ordered triples \((a,b,c)\) is \(a^2b+b^2c+c^2a-ab^2-bc^2-ca^2 = 0\)?
I wrote a computer program and got 80 triples:
for (a = 1..5)
for (b = 1..5)
for (c = 1..5)
if a^2*b + b^2*c + c^2*a - a*b^2 - b*c^2 - c*a^2 = 0
count = count + 1
output(count)
output = 80
+140 
