De Viete gives us $ 2a+b+c = 0$, $ c =-abc$ and $ b = ab+bc+ca$. If $ c = 0$: if also $ b = 0$ then $ f(x) = x^{3}$, else $ a = 1$, so $ b =-2$, and $ f(x) = x^{3}+x^{2}-2x$ is ok. If $ c\not = 0$ then $ ab =-1$ and thus $ a,b\not = 0$. Mutliply the last equation with $ a$ to get $ ab = a^{2}b+abc+ca^{2}$ or $ -1 =-a+c+ca^{2}$, yielding $ c =\frac{a-1}{a^{2}+1}$. On the other hand is $ a(-2a-c) =-1$, so $ c =\frac{1-2a^{2}}{a}$. Since $ a(a-1) = (1+a^{2})(1-2a^{2})\Leftrightarrow 1+a-2a^{2}-2a^{4}= 0$ has no rational roots, no solutions here. So $ f(x)\in\{x^{3},x^{3}+x^{2}-2x\}$.

You are missing the solution a=-b=-c=1