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Since $a,b,c$ are the roots of the cubic polynomial, Vieta's formulas give us:

$$\begin{array}{rl}-(a+b+c)& =a\\ ab+bc+ca& =b\\ -abc& =c\end{array}\text{\hspace{1em}(1)(2)(3)}$$Let's do this with casework. Assume $c=0.$ This satisfies equation (3). Equation (1) translates to $2a+b=0,$ and equation (2) translates to $ab=b.$ If $b=0,$ then $a=0.$ If $b\ne 0,$ then $a=1$ and $b=-2.$

Now assume $c\ne 0.$ Equation (3) then requires that

$$\begin{array}{r}ab=-1.\end{array}\text{\hspace{1em}(4)}$$Equation (2) then becomes $-1+c(a+b)=b.$

Let $a+b=0.$ Then (2) gives $b=-1,a=1,$ and (1) then gives $c=-1.$ This is our third solution.

If $c\ne 0$ and $a+b\ne 0,$ then from the equation $-1+c(a+b)=b$,

$$c=\frac{b+1}{a+b}=\frac{a(b+1)}{a(a+b)}$$Using (4) to simplify:

$$c=\frac{-1+a}{a^2-1}=\frac{1}{a+1}$$Now (1) gives

$$-\left(a-\frac{1}{a}+\frac{1}{a+1}\right)=a.$$Or $2a^3+2a^2-1=0.$ However, this has no rational roots (we can test $a=\pm 1,\pm 1/2$). Therefore we have $\boxed{3}$ solutions: $(0,0,0)$, $(1,-2,0)$, and $(1,-1,-1)$.