Winning strategy for a game with cubic equation The following problem is from USSR $1990$:

The following equation with erased coefficients is written on a blackboard: $$x^3+\dots x^2+\dots x+\dots =0$$
Two players are playing a game. In one move the first player chooses a number and the second player puts it instead of dots into one of the vacant places. After three moves the game is over. Is it possible for the first player to choose three numbers that will secure three distinct integer roots for the equation, no matter how the second player plays?

The solution to this problem which I found from a book is as follows:
Solution from book: Yes, it is possible. One strategy is as follows. At the beginning the first player chooses $0$.
$\color\green{\text{Case}\ 1}$: If the second player makes it the constant term of the equation, the equation will be $$x^3+\dots x^2+\dots x=0$$ and the first player chooses successively $2$ and $-3$ to obtain $$x(x-1)(x+3)=0\ \ \text{or}\ \ x(x-1)(x-2)=0$$
$\color\red{\text{Case}\ 2}$: If the second player puts $0$ as the coefficient of $x^2$, then the equation becomes $x^3+bx+c=0$ with $b$ and $c$ not fixed yet. The first player chooses the number $\color\red{-(3\cdot4\cdot5)^2}$ and then depending on the move of the second player, either $c=0$ or $b=3^2\cdot4^2-4^2\cdot5^2-3^2\cdot5^2$. This will result in the equations $$x(x-3\cdot4\cdot5)(x+3\cdot4\cdot5)=0$$ or $$(x+3^2)(x+4^2)(x-5^2)=0$$
$\color\red{\text{Case}\ 3}$: If after the move of the second player the equation is $x^3+ax^2+c=0$, the first player chooses $\color\red{6^2\cdot7^3}$ and then either $c=-49$ or $c=-6^8\cdot7^6$ to get the equations $$(x+2\cdot7)(x-3\cdot7)(x-6\cdot7)=0$$ or $$(x-2\cdot6^2\cdot7^2)(x+3\cdot6^2\cdot7^2)(x+6\cdot6^2\cdot7^2)=0$$
In all cases, we got equations with three distinct integer roots and we are done.$\ \ \blacksquare$
I understand the case 1 of the solution. But I don't understand the case 2 and case 3 (which is in red) of the solution, especially how the numbers in the second move (numbers in red) are chosen in these two cases. I know I have to use Vieta's formula. I have seen that choosing the numbers in the second move arbitrarily doesn't work.

In case 2, the idea is:

*

*If $N$ is a perfect square, then $x^3 - Nx$ factors as $x(x+\sqrt N)(x-\sqrt N)$. This means we can win by playing $-N$, then $0$ if $-N$ is chosen as the coefficient of $x$.

*If $N = pq(p+q)$, then the expansion of $(x+p)(x+q)(x-p-q)$ has an $x^2$ coefficient of $0$ and constant term of $-N$. This means that we can win by playing $-N$ if it is chosen as the constant term: our next play is whatever the coefficient of $x$ in $(x+p)(x+q)(x-p-q)$ is.

The easiest way to make sure $pq(p+q)$ is a perfect square is to make $p$, $q$, and $p+q$ all perfect squares. (That's not the only way: $2 \cdot 16 \cdot (2+16)$ is a perfect square, for example. But such solutions are harder to find.) And the easiest way to make $p$, $q$, and $p+q$ all perfect squares is to set $p=3^2$ and $q=4^2$ so that $p+q = 5^2$.

Case 3 is the hardest, because our three roots $p,q,r$ must satisfy $pq+pr+qr=0$, which is not a terribly nice condition. In both subcases, the idea we are inspired by is that $(x+2)(x-3)(x-6)$ has this property (which we can check) and therefore $(x+2k)(x-3k)(x-6k)$ also has this property (because this scales all three terms of $pq+pr+qr$ by $k^2$).
The expansion of $(x+2k)(x-3k)(x-6k)$ is $x^3 - 7kx^2 + 36k^3$. So to force a polynomial of this form we need a value of $N$ that is

*

*of the form $-7k_1$ for some $k_1$, and

*of the form $36k_2^3$ for some $k_2$.

The most straightforward way to satisfy both conditions is to set $N = 36 \cdot 7^3$.

*

*If $N$ is chosen as the coefficient of $x^2$, then we can get the polynomial $(x+2k)(x-3k)(x-6k)$ with $k = -36 \cdot 7^2$.

*If $N$ is chosen as the constant term, then we can get the polynomial $(x+2k)(x-3k)(x-6k)$ with $k = 7$.

