x = a/3 y = b/5 z = - c/6
a + b + c = 2
5 (a/3)(b/5) - 10(b/5)(-c/6) - 6(a/3)(-c/6) = -41
abc / -90 = 6
a + b + c = 2 ⇒ a + b = 2 - c (1)
ab + bc + ac = -123 ⇒ ab + c ( a + b) = -123 (2)
abc = -540 ⇒ ab = -540/c (3)
Sub (1) and (3) into (2)
-540/ c + c (2 - c) = -123
-540/c + 2c - c^2 = -123
-540 + 2c^2 - c^3 = -123c
c^3 - 2c^2 - 123c + 540 = 0 re-write as
c^3 - 2c^2 - 63c - 60c + 540 = 0
c(c^2 - 2c - 63) - 60c + 540 = 0
c(c -9)(c + 7) - 60 (c - 9) = 0
(c - 9) [ c(c + 7) - 60 ] = 0
(c - 9) [ c^2 + 7c - 60] = 0
(c - 9) ( c + 12)(c - 5) = 0
So...possible values of c = 5, 9 and - 12
z is an integer if c = -12
And ab = 45
And a + b = c - (- 12) ⇒ a + b = 14
And x is an ineger if a = 9
And y is an integer if b = 5
So the cubic monomial we need is
(t + 12) (t -9) (t - 5) =
t^3 - 2 t^2 - 123 t + 540
