Common complex roots 
If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.

My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $(x^2+x+1)=0$ and $ax^2+bx+c=0$ are common should imply $a=b=c$ not only this but $a=b=c=1$.
Is this solution correct?

We have that
$$
\eqalign{
  & x^{\,3}  + 3x^{\,2}  + 3x + 2 = 0\quad  \Rightarrow \quad x^{\,3}  + 3x^{\,2}  + 3x + 1 =  - 1\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {x + 1} \right)^{\,3}  =  - 1\quad  \Rightarrow \quad \left( {x + 1} \right) = e^{\,i\;\,\left( {1 + 2k} \right)\pi /3}  = e^{\, \pm \,i\;\,\pi /3} ,e^{\,i\;\pi }  = {1 \over 2} \pm i{{\sqrt 3 } \over 2},\;\; - 1 \cr} 
$$
and
$$
\eqalign{
  & 0 = ax^2  + bx + c\quad  \Rightarrow \quad a\left( {x + 1} \right)^2  + \left( {b - 2a} \right)\left( {x + 1} \right) + \left( {a + c - b} \right) = 0\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {x + 1} \right) = {{ - \left( {b - 2a} \right) \pm \sqrt {b^2  - 4ac} } \over {2a}} \cr} 
$$
Comparing the two results, the condition to have two identical roots imposes that it shall be
$$
\left\{ \matrix{
  {{2a - b} \over {2a}} = {1 \over 2} \hfill \cr 
  {{\sqrt {4ac - b^2 } } \over {2a}} = {{\sqrt 3 } \over 2} \hfill \cr}  \right.\;\quad  \Rightarrow \quad \left\{ \matrix{
  b = a \hfill \cr 
  c = a \hfill \cr}  \right.
$$
either in the real and in the complex field.
