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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.

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[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.

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    @Raj: Please note that the first version (to which your comment refers) contained$a$mistake that caused one solution go missing; sorry about that.2012-12-07