Q:

Find polynomial : $P(x)=x^3+ax^2+bx+c$

Find all polynomials in $\mathbb{Q}[x]$ is of the form $P(x)=x^3+ax^2+bx+c$ which has $a, b, c$ as its roots.
Is my answer below correct?

A:

$x^3+ax^2+bx+c = (x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$
so $a=-(a+b+c)$, $b=ab+bc+ca$, $c=-abc$
Case 1 : $c=0$ so $b=ab$ then $a=1$ or $b=0$
if $a=1$, we get $(a,b,c)=(1,-2,0)$
if $b=0$, we get $(a,b,c)=(0,0,0)$
Case 2 : $c\not=0$ so $ab=-1$ then $-1+c(a+b)=b$ ---[1]
Since $a=-(a+b+c)$, so $c=-2a+\frac{1}{a}$ 
substitute $c=-2a+\frac{1}{a}$ and $b=-\frac{1}{a}$ in [1]
we have $1+(a-\frac{1}{a})(2a-\frac{1}{a})=\frac{1}{a}$, then $(a-1)(2a^2(a+1)-1)=0$
If $2a^2(a+1)-1=0$, then $2a^3+2a^2-1=0$, by Rational root theorem, a $\not \in \mathbb{Q}$
so $a=1$, we get $(a,b,c)=(1,-1,-1)$
Answer : $P(x)=x^3$, $x^3+x^2-2x$, $x^3+x^2-x-1$

A:

By Vieta's formulas,
$$a+b+c=-a,\\ab+bc+ca=b,\\abc=-c.$$
Then if $c\ne 0$,
$$ab=-1$$
$$c=-2a-b$$
$$-1-(a+b)(2a+b)=-2a^2+2-b^2=b$$ and
$$2a^2b^2=(2-b-b^2)b^2=1.$$
The last equation has no rational roots and the only remaining option is $c=0$, giving
$$2a+b=0,ab=b$$
with the two solutions $a=b=0$, and $a=1,b=-2$.

