Another cubic, with interesting roots

It can be easily shown that

$r_1r_2r_3 + (r_1 + r_2)(r_2 + r_3)(r_3 + r_1) = (r_1 + r_2 + r_3)(r_1r_2 + r_2r_3 + r_3r_1).$

(See this note .)

Since $-a = r_1 + r_2 + r_3$ , $b = r_1r_2 + r_2r_3 + r_3r_1$ , and $-a = r_1r_2r_3$ , we have

$\begin{aligned} -a - 1 &= -ab \\ -1 &= -ab + a \\ 1 &= a(b - 1). \end{aligned}$

Since $a$ and $b$ are integers, we have either $a = 1$ and $b - 1 = 1$ , or $a = -1$ and $b - 1 = -1.$ The second case gives $b = 0,$ which is not allowed. Thus, $b = 1 + 1 = \boxed{2},$

We have $r_1 + r_2 + r_3 = -a$

$\Rightarrow (r_1 + r_2)(r_2 + r_3)(r_3 + r_1) = (-a - r_3)(-a - r_1)(-a - r_2)$

$= f(-a) = -ab + a = -1 \Rightarrow a(b-1) = 1$

Hence we have either $a=1$ and $b=2$ or $a=-1$ and $b=0$ . But since $b$ is nonzero integer we have $b=2$

$(r_1\times r_2 + r_1\times r_3 + r_2\times r_3 + r_2^{2})\times (r_3 + r1) = -1$ or $\ r_1\times r_2 + r1\times r_3 + r_2\times r_3 = b$ and $r_3+r1 = -a -r_2$
then $\\ -(b+r_2^{2})(a+r_2) = -1 \Leftrightarrow r_2^{3} + r_2^{2}a + br_2 + ab = 1$ $\Leftrightarrow r_2^{3} + r_2^{2}a + br_2 + ab + a = 1 + a$ or $r_2$ is a root then $r_2^{3} + r_2^{2}a + br_2 + ab + a = 0\$ then we have $ab = 1+a \Leftrightarrow b = 1 +1/a$ $a$ and $b$ are non zero integers : $\boxed{a = 1} \ \boxed{ b = 2}$