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Let \(0 \le a, b, c \le 5\)  be integers. For how many ordered triples \((a,b,c)\)  is \(a^2b+b^2c+c^2a-ab^2-bc^2-ca^2 = 0\)?

 Aug 20, 2021
 #1
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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

 Aug 20, 2021

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