let f(x) = ax^3 + bx^2 +c x +d
f(x) - 12 can have at most 3 distinctive solutions
and f(x) + 12 is also.
since |f（1）| = |f(2)| = |f(3)| = |f(5）| = |f(6)| = |f（7_)| = 12, without loss generality, we assume at least of three from , 1, 2, 3, 5, 6, 7 will have f(x) = 12. Since f(x) - 12 can have at most three real solution, so it must be exact three of them. In other words, other three will have f(x) = -12.
if we assume f(7) = 12, then f(5) = f(6) = -12, and f(2) = f(3) = 12, and f(1) = -12 from 3rd order polynomial properties.
so f(x) - 12 = 0 has solution 2, 3, 7, and f(x) + 12 has solutions of 1, 5, 6.
Hence we have (d -12)/a = -2*3*7 (from f(x) - 12 = 0)
(d + 12)/a = -1 * 5 * 6
(d - 12)/(d + 12) = 7/5
solve for d = -72. certainly d = 72 is also a solution f(0) = d
