(0) Assume there is a solution other than A = B = C = 0
(1) Clearly, if two of them are zero, the 3rd must be 0 also.
(2) Further, assume one of them is zero and the other two are not. Then either (A^3 = -3B^3) or (A^3 = -9C^3) or (B^3 = -3C^3)
Clearly, none of these have a rational solution, because 3 and 9 have irrational cube roots. Therefore, we have a contradiction, so assumption (2) is incorrect.
(3) So, we know that A, B and C are all non-zero. Express them as fractions in lowest terms.
Multiply A, B and C by the least common multiple (LCM) of their denominators, giving D, E and F. Now D, E and F are integers.
Divide D, E and F by their greatest common factor (GCF), giving G, H, and I. G, H and I are integers without a factor common to all three of them.
(4) It must the case that G^3 + 3H^3 + 9I^3 = 3GHI.
But this means that G must be a multiple of 3, because the other terms are all multiples of 3.
Let G = 3J and substitute.
Then 27J^3 + 3H^3 + 9I^3 = 9JHI.
Dividing by 3 gives 9J^3 + H^3 + 3I^3 = 3JHI.
But this means that H must be a multiple of 3, because the other terms are all multiples of 3.
Let H = 3K and substitute.
Then 9J^3 + 27K^3 + 3I^3 = 9JKI.
Dividing by 3, 3J^3 + 9K^3 + I^3 = 3JKI.
Now I must be a multiple of 3, because the other terms are all multiples of 3.
(5) And at this point, we have reached a contradiction. G, H and I are all multiples of 3. But this is impossible, because we got them by dividing D, E and F by their greatest common factor.
Therefore, assumption (0) is false. The only solution is A = B = C = 0.
q.e.d.
q.e.d. (spoiler)
