If $a,b,c$ are rational roots for $x^3+a x^2+b x+c=0$, find the value of a, b, c 
Given: $a,b,c$ are rational roots for the equation $x^3+a x^2+b x+c=0$
Find: all possible values of $a,b,c$.

Question used within the preparation for an entrance exam.
My attempt: from Vieta's formulas, we know that (1) $a+b+c=-a$, (2) $ab+bc+ac=b$ and (3) $abc=-c$. Developing this last equation, we get $c(ab+1)=0$. Let's assume that $c=0$. Replacing $c=0$ in (2) we get $ab=b$, or $a=1$. Replacing $a=1$, $c=0$ in (1) we get $b=-2$. Therefore $a=1,b=-2,c=0$ is a possible set of values. By inspection $a=b=c=0$ is another set.
Questions: are there other (rational) values for $a,b,c$ beyond these 2 ones I found? Are there other approaches to address this question?
 A: You already have $c(ab+1)=0$. But you only considered one case $c=0$.
Let consider the case $ab=-1$. One then has, from (2), $c(a+b) = b+1$
Or $c(a^2-1)=a-1$.
Subcase 1: $a=1$, then $b=-1$. From (1), one has $c=-2a-b=-1$.
Subcase 2: $c(a+1)=1$, then from (1), one has $$a-\frac{1}{a} + \frac{1}{a+1}=-a$$, then $$2a-\frac{1}{a(a+1)}=0$$ $$2a^2(a+1)=1.$$ Solve this equation, you don't have $a$ is rational (for example, link).
So you have one more tuple $(1,-1,-1)$.
A: The way the question is phrased, it doesn't say that $a$, $b$, $c$ have to be all the roots of the equation (or in case some of them are equal, as in $(0,0,0)$, that they will list the roots with the correct multiplicities). In other words, Viet doesn't have to apply as $abc = -c$, it could be $aac = -c$ (if, for example, $a$ is a double root, while $b = c$ is a single root), or even $abd = -c$, where $d$ is another root (not one of $a$, $b$, $c$) while $a = c \neq b$ are two single roots.
Because of this, when starting with Viet, you might miss a solution. In fact you both (bluemaster and GAVD) have: it's $(-1, -1, 1)$. The roots are then $\{-1, 1\}$ but the double root is the $1$, not the $-1$. That's why the Viet approach misses it.

A more general approach is to use the theorem about rational roots to first narrow things down, then use the definition of what a root is to exhaust the possibilities. The theorem says that a rational root must be of the form $p/q$, where $p$ is a divisor of the constant term, while $q$ is a divisor oh the leading coefficient. In our case, this means $q = 1$ (so $a$, $b$, $c$ are all integers), and $p$, hence each of $a$, $b$, $c$, is a divisor of $c$.
You can now distinguish two cases:


*

*If $c=0$, after factoring out an $x$ we get a quadratic $x^2 + ax + b$ which must have both $a$ and $b$ as roots. Thus $2a^2 + b = 0$ and $b^2 + ab + b = 0$ which easily gives two integer solutions, $(0,  0)$ and $(1, -2)$.

*If $c \neq 0$, then since $c$ is a root, $c^3 + ac^2 + (b+1)c = 0$. The first two terms are divisible by $c^2$, so also must be the third one. This means $c$ is a divisor of $b+1$, and thus $b$ (which is a divisor of $c$) divides $b+1$. This is only possible if $b = 1$ or $b = -1$. Now consider two subcases, proceeding as in the previous case to make simultaneous equations for $a$ and $c$. You get the other two possibilities, $(1, -1, -1)$ and $(-1, -1, 1)$.
