The roots of the cubic are 5, -12, and 9, and you would find (x, y, z), and from that x+y, by assigning a permutation of the roots to (3x, 5y, -6z). Three numbers can be permuted in 6 different ways, so there are 6 different assignments:
(3x, 5y, - 6z)=(9, 5, -12)
=(9, -12, 5)
= (5, 9, -12)
=(5, -12, 9)
= (-12, 5, 9)
=(-12, 9, 5)
You would find 6 values for x +y this way, and I think they turn out to be distinct.
