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?

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 $a=1,b=-2$.
