A "combinatorial" approach to a problem about the roots of a polynomial

Consider the following problem. 
Problem (4b - Algebra for Group A or 5b - Algebra for Group B, National Student Mathematical Olympiad): Consider the polynomial T(x) = x^3 - 3x. Prove that there doesn't exist any polynomial S(x) of degree 3 such that S(T(x)) has 9 distinct roots that form an arithmetic progression. 
This problem can be solved combinatorially as follows. 
Assume that there is a such polynomial S(x). Then, S(x) must have three distinct roots a, b, c and the 9 distinct roots of S(T(x)) consist of 3 roots of x^3 - 3x - a, 3 roots of x^3 - 3x - b and 3 roots of x^3 - 3x - c. Let the nine roots of S(T(x)) be A, A + d, A + 2d, ..., A + 8d, where d is a nonzero real number. Assume that: 
   i) The three roots of x^3 - 3x - a are A + a(1)d, A + a(2)d, A + a(3)d,  
  ii) Similarly, the three roots of x^3 - 3x - b are A + b(1)d, A + b(2)d, A + b(3)d and the three roots of x^3 - 3x - c are A + c(1)d, A + c(2)d, A + c(3)d. 
Moreover, a(1), a(2), a(3), b(1), b(2), b(3), c(1), c(2), c(3) are the eight numbers 0, 1, 2, ..., 8 in some order. 
By Vieta's theorem, (A + a(1)d) + (A + a(2)d) + (A + a(3)d) = 0 = 3A + (a(1) + a(2) + a(3))d and a(1) + a(2) + a(3) = -3A/d. Similarly, we also have b(1) + b(2) + b(3) = c(1) + c(2) + c(3) = -3A/d. As a result, a(1) + a(2) + a(3) = b(1) + b(2) + b(3) = c(1) + c(2) + c(3) = (0 + 1 + 2 + ... + 8)/3 = 36/3 = 12. 
Again, by Vieta's theorem, (A + a(1)d)(A + a(2)d) + (A + a(2)d)(A + a(3)d) + (A + a(1)d)(A + a(3)d) = -3 = 3A^2 + 2Ad(a(1) + a(2) + a(3)) + d(a(1)a(2) + a(2)a(3)+a(1)a(3)), thus a(1)a(2) + a(2)a(3) + a(1)a(3) = (-3+6A^2)/d. Similarly, we also have b(1)b(2) + b(2)b(3) + b(1)b(3) = c(1)c(2) + c(2)c(3) + c(1)c(3) = (-3+6A^2)/d. Therefore, a(1)^2 + a(2)^2 + a(3)^2 = b(1)^2 + b(2)^2 + b(3)^2 = c(1)^2 + c(2)^2 + c(3)^2 = (0^2 + 1^2 + ... + 8^2)/3 = 204/3 = 68. 
Without loss of generosity, assume that a(1) = 8. Then, a(2) + a(3) = 4 and a(2)^2 + a(3)^2 >= 2^2 + 2^2, thus a(1)^2 + a(2)^2 + a(3)^2 = 8^2 + 2^2 + 2^2 > 68, a contradiction. This contradiction shows that a such polynomial S(x) doesn't exist. 
Comment: The combinatorial part of this problem can be stated as follows: It is possible to partition A = {0, 1, 2, ..., 8} into three sets, each of cardinality three, such that the sum of the elements in these three sets are equal and the sum of the squares of the elements in these three sets are also equal? Since 8^2 = 64 is quite large in comparison to other squares, this part is quite simple as soon as we realize it. Moreover, T(x) = x^3 - 3x can be replaced by any other polynomial, since we only need the sum of the three roots and the sum of the squares of the three roots to be fixed. 

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