+0 Find the number of ordered triples (a,b,c) of integers such that 0≤a,b,c≤5 and
a^2b+b^2c+c^2a-ab^2-bc^2-ca^2=0
+195 Factoring the equation, we get a2(b−c)+b(b2−c2)+c(c2−a2)=0. This can be rewritten as (b−c)(a2−ab+c)+b(b−c)(b+c)=0.
Therefore, (b−c)(a−b)(a−c)+b(b−c)(b+c)=(b−c)(a−b)(a−c)(b+c)=0. So b−c or a−b, a−c, or b+c must equal 0.
If b−c=0, then b=c. Since 0≤b,c≤5, it follows that b=c=0,1,2,3,4,5. In each case, we can choose a to be any integer between 0 and 5, inclusive.
This gives us 6⋅6=36 possibilities.
Therefore, the answer is 36.
