Find all rational triplets $(a,b,c)$ that are roots of the equation $$x^3+ax^2+bx+c=0$$
My work so far:
By Vieta's formulas we have
$a+b+c=-a\;(1),$
$ab+ac+bc=b\;(2),$ and
$abc=-c\;(3).$
From $(3),$ either $c=0$ or $ab=-1$. If $c=0,$ then $(1)$ becomes $b=-2a,$ and $(2)$ becomes $b(1-a)=0$. Hence $a=b=0$ or $a=1\Rightarrow b=-2.$ So assume $c\neq 0$ and $ab=-1.$ $(1)$ becomes $c=-b-2a.$ Substituting in $(2)$ we have $-1-ab-2a^2-b^2-2ab=b\Rightarrow-1-(2a+b)(a+b)=b,$ so $-a^2-2a^4+3a^2-1=-a,$ or $2a^4-2a^2-a+1=0.$ So $a=1$ or $2a^3+2a^2-1=0.$ The first possibility gives $b=-1$ and $c=-1.$ Suppose $m/n$ is a root of $2a^3+2a^2-1=0$ with relatively prime integers. Then $2m^3/n^3+2m^2/n^2-1=0\Rightarrow 2m^3+2m^2n-n^3=0.$ So any prime factor of $n$ must divide $2$ and any prime factor of $m$ must divide $1.$ Hence the only possibilities are $-1/2,1/2,\pm 1$ and it is easy to check that they are not solutions. Thus, the only rational triplets are $\boxed{(0,0,0),(1,-2,0),(1,-1,-1)}.$
Two years too late
$$(x^3+ax^2+bx+c)-(x-a)(x-b)(x-c)=$$ $$c(a b +1)- (c (a+b)+b(a-1))x+ (2 a+b+c)x^2=0\tag 1$$ So $c=-2a-b$ replaced in $(1)$ leads to $$-(2 a+b) (a b+1)+[2 a (a+b)+b (b+1)]x=0 \tag 2$$
If $\color{red}{b=-2a}$, then $a(a-1)=0$, that is to say $a=0$ or $a=1$.
If $a=0$, then $b=0$ and $c=0$
If $a=1$, then $b=-2$ and $c=0$
If $\color{red}{b=-\frac 1a}$, we have $$\frac{\left(2 a^4-2 a^2-a+1\right) }{a^2}=\frac{(a-1) \left(2 a^3+2 a^2-1\right)}{a^2}=0$$
So, if $a=1$, $b=-1$ and $c=-1$
The last case is $(2a^3+2a^2-1)=0$, the real solution of which being $$\frac{1}{3} \left(2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{23}{4}\right)\right)-1\right)$$ which has very little chances to be rational.
Then, your solution and conclusions just using algebra.