Polynomial game problem: do we have winning strategy for this game? I'm thinking about some game theory problem. Here it is,

Problem: Consider the polynomial equation $x^3+Ax^2+Bx+C=0$. A priori, $A$,$B$ and $C$ are "undecided", yet and two players "Boy" and "Girl" are playing game in following way:
  
  
*
  
*First, the "Boy" chooses a real number.
  
*Then, the "Girl" decides the position of that number among $A$,$B$ and $C$.
  
*They repeat this process three times, deciding all values for $A$,$B$ and $C$.
  
  
  If the final polynomial has all distinct integer roots, the Boy wins. Otherwise, the Girl wins. 
Question is: Does one of Boy or Girl has any "Winning Strategy"? If so, who has one?

It seems to me, obviously the boy has great advantage. Here is my attempted argument: If the boy suggest "$0$" at the very first turn, regardless of the girl's decision we can make the polynomial to have three distinct integer roots. Actually, my argument has "almost" worked. for example,


*

*If the girl put $0$ at position $A$, then boy should suggest "$-84$" at the second round. Then, in any case, we always have three distinct roots, i.e. $(10,-2,-8)$ or $(-3,-4,7)$.

*If the girl put $0$ at position $C$, then boy should suggest "$-1$" at the second round. Then, in any case, we always have three distinct roots, i.e. $(-1,2,0)$ or $(1,-1,0)$.

*HOWEVER, if the girl put $0$ at position $B$, I couldn't find any good number for second round.


Has my winning strategy some problem? Or has the girl a winning strategy somehow?
Thank you for any help in advance!

If the girl puts $0$ at position $B$, the boy can choose $1764$, resulting in roots $[-6, 7, 42]$ or $[-864, -1440, 540]$.
EDIT:  If the polynomial $(x-a)(x-b)(x-c)$ has $ab+bc+ac=0$, then $c = -ab/(a+b)$.  Writing $t = a+b$, we need $t$ to divide $ab = at - a^2$.  Thus $t$ is a divisor of $a^2$.
Here's a Maple program that finds the solution $1764$, together with some others.  For positive integer values of $a$, it looks at each divisor $t$ of $a^2$ (positive as well as negative), computes the corresponding $b$ and $c$, and the corresponding values of the coefficients $A$ and $C$ of the polynomial, storing under those indices the values $[a,b,c]$.  We then look for an index that appears for both $A$ and $C$.
for a from 1 to 1000 do
  for t in numtheory:-divisors(a^2) union map(`*`,numtheory:-divisors(a^2),-1) do
     b:= t-a;
     c:= -a*b/t;
     if nops({a,b,c}) < 3 then next fi;
     A[-a-b-c]:= [a,b,c];
     C[-a*b*c]:= [a,b,c];
  od
od:
{indices(A)} intersect {indices(C)};

$$ \{[-12348], [-1764], [1764], [3136], [12348]\}$$
A[1764], C[1764];

$$[42, 7, -6], [540, -864, -1440]$$
