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?

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 $(1,-1,-1)$.
