I'll look at one of yours
\(a(ab+c^2) + b^2c = ab^2 + bc^2 + ca^2\\ a(ab+c^2) + b^2c = a(b^2+ca) + bc^2 \\ \text{I will assume a can equal anything at all}\\ \text{This means that}\\ ab+c^2=ca+b^2 \qquad and \qquad b^2c=bc^2\)
this would always be true if b=c
given the restiction that they all must be between 0 and 5 inclusive,
If a=0 these are all possibilities
a=0, b=0,c=0
a=0, b=1,c=1
a=0, b=2,c=2
a=0, b=3,c=3
a=0, b=4,c=4
a=0, b=5,c=5
If a=1 these are all possibilities
a=1, b=0,c=0
a=1, b=1,c=1
a=1, b=2,c=2
a=1, b=3,c=3
a=1, b=4,c=4
a=1, b=5,c=5
etc a can be 0,1,2,3,4, or 5 that is 6 possible numbers for a, each of these have 6 possibilities for b=c
That makes 6*6=36 possibilities here
Now do the other 2 factorisations, what do you get?
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