$x^3+a{x}^2+bx+c=0=(x-a)(x-b)(x-c)$

$a+b+c=-a$

$\Rightarrow 2a+b+c=0 \dots \left(\mathrm{i}\right)$

$ab+bc+ca=b \dots \left(\mathrm{ii}\right)$

$abc=-c\Rightarrow ab=-1[\because c\ne 0] \dots \left(\mathrm{iii}\right)$

Also a is a root of equation

$\Rightarrow 2a^3+ab+c=0\Rightarrow 2a^3-1+c=0$

$\Rightarrow c=1-2a^3$

from $\left(\mathrm{i}\right)$

$2a^2+ab+ac=0$

$2a^2-1+a\left(1-2a^3\right)=0$

$2a^2-2a^4+a-1=0$

$2a^2(1-a)(1+a)+(a-1)=0$

$\Rightarrow (1-a)\left[2a^2(a+1)-1\right]=0$

$\Rightarrow a=1$ or $2a^3+2a^2-1=0$

when $a=1, b=\frac{-1}{a}=-1$ and $c=1-2a^3=-1$

when $2a^3+2a^2-1=0$

There will be only one real solution of

$f\left(x\right)=2x^3+2x^2-1=0$

as $f^{\text{'}}\left(x\right)=6x^2+4x=0\Rightarrow x=0,\frac{-2}{3}$

$f\left(0\right)·f\left(\frac{-2}{3}\right)<0$

$\therefore$ corresponding to this real value of a one triplet

is possible

$\therefore$ Exactly two triplets $(a,b,c)$ are possible