I presume that the 3's are exponents, and a, b and c must be positive integers.
So you are looking for a cube that's equal to the sum of two other cubes.
If the exponent were 2, these would be "Pythagorean Triples," which could represent the 3 sides of a right triangle.
Finding solutions (or a proof that there aren't any), for exponents 3 or greater is known as "Fermat's Last Problem" because he claimed to have a clever proof that there were no solutions. Today, mathematicians believe that he did NOT in fact have a proof. An extremely long proof using very advanced techniques finally did prove that there are no solutions. Wikipedia says "It is among the most famous theorems in the history of mathematics and prior to its 1995 proof was in the Guinness Book of World Records for "most difficult mathematical problems".
You dad is safe.
You can get back at him. Offer HIM something valuable if he can draw the following diagram with one continuous line, without lifting the pencil from the paper or retracing a line. This can't be done (consider where you would have to start and end--each vertex has 5 lines coming into it, but you can only start and stop at 2 vertices, so this takes two lines.)
http://i276.photobucket.com/albums/kk2/freond1/drawproblem.png
It's a square, with both diagonals drawn in, and with an arc between adjacent vertices. I.e., a square inscribed in a circle, with diagonals too.
You can find more info at http://en.wikipedia.org/wiki/Fermat%27s_last_theorem