$f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$?
The polynomial is allowed to have repeated roots.
 A: [There was a mistake in the first version of this answer that caused one solution to go missing.]
The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have
$$
\begin{align}
a&=-a-b-c\;,\\
b&=ab+bc+ca\;,\\
c&=-abc\;.
\end{align}
$$
If $c=0$, this becomes
$$
\begin{align}
a&=-a-b\;,\\
b&=ab\;,\\
\end{align}
$$
and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$.
If $c\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields
$$
c=-2a+\frac1a\;,
$$
and then the second equation becomes
$$
-\frac1a=-1+2-\frac1{a^2}-2a^2+1\;.
$$
Multiplying through by $a^2$ yields
$$
2a^4-2a^2-a+1=0\;.
$$
The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields
$$
2a^3+2a^2-1=0\;,
$$
which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$.
Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively.
