Symmetree

Multiplying the third equation by $abc$ we get that $ab+bc+ca = 3$ . Then by Vieta's, $a,b,c$ are the roots of the equation $x^3+3x^2+3x+1 =0$ which can be factored as $(x+1)^3$ . Hence, $(a,b,c) = (-1,-1,-1)$ is the only solution.

It is obvious that $(a, b, c)=(-1,-1,-1)$ is an answer. Let us proof that there exist this answer only.

By the third equation, we can get $ab+ac+bc=3$ . Now, square the first equation, we will get $\begin{aligned} a^2+b^2+c^2+2ab+2bc+2ac&=9 \\ a^2+b^2+c^2&=3 \end{aligned}$

Solution 1) Now apply Cauchy inequaliy, $(a^2+b^2+c^2)(1^2+1^2+1^2) \geq (a+b+c)^2$ $LHS=3\times3=9$ $RHS=(-3)^2=9$ Which means that Cauchy equality hold. Hence, a=b=c. Combine it with the first equation, we can conclude that there exist only $\large 1$ triple, which is $(-1,-1,-1)$ .

Solution 2) We got $\begin{aligned} a+b+c&=-3 \\ 2a+2b+2c&=-6\\ a^2+b^2+c^2&=3 \\ a^2+2a+b^2+2b+c^2+2c&=-3\\ (a+1)^2+(b+1)^2+(c+1)^2&=0\\ \end{aligned}$ Hence, $a+1=b+1=c+1=0$ $a=b=c=-1$

$\begin{aligned} \frac 1a + \frac 1b + \frac 1c & = -3 & \small \color{#3D99F6} \text{Given} \\ \frac {ab+bc+ca}{\color{#3D99F6}abc} & = - 3 & \small \color{#3D99F6} \text{Given that }abc = -1 \\ \frac {ab+bc+ca}{\color{#3D99F6}-1} & = - 3 \\ ab+bc+ca & = 3 & \small \color{#3D99F6} \text{Multiply both sides by }a \\ a^2b+{\color{#3D99F6}abc} + ca^2 & = 3a \\ a^2b{\color{#3D99F6}-1} + ca^2 & = 3a \\ a^2(b+c) & = 3a+1 \\ a^2({\color{#3D99F6}a + b + c} - a) & = 3a+1 & \small \color{#3D99F6} \text{Given that }a+b+c = -3 \\ a^2({\color{#3D99F6}-3} - a) & = 3a+1 \\ a^3+3a^2+3a+1 & = 0 \\ (a+1)^3 & = 0 \\ \implies a & = -1 \end{aligned}$

Since $a$ , $b$ and $c$ are identical in the system of equations, there is only $\boxed{1}$ triple $(-1,-1,-1)$ satisfies the system.