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?

If $a$, $b$, and $c$ are not real, then the question as stated is false.  The roots of $x^3 + 3x^2 + 3x + 2$ are $-2$ and $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$;  and the polynomial $x^2 + (\frac{5}{2} + \frac{\sqrt{3}}{2} i)x + (1 + \sqrt{3} i) = (x + 2)(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)$ shares two roots with $x^3 + 3x^2 + 3x + 2$ but does not have $a = b = c$.  
I think that the strongest statement that can be made is that if $a, b, c \in \mathbb{C}$ have the same argument, then the statement must be true.  If this is the case, then we have $a = Ae^{i\theta}, b = B e^{i\theta}, c = e^{i \theta}$ for the same $\theta$ in each case and $A, B, C \in \mathbb{R}$.  In such a case, the polynomial $a x^2 + bx + c = e^{i \theta}(A x^2 + Bx + C)$ will have the same roots as $Ax^2 + Bx + C$;  and since this latter polynomial is real, its roots must be complex conjugates of each other.  This implies that $A x^2 + Bx + C$ is a multiple of $(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)(x +  \frac{1}{2} - \frac{\sqrt{3}}{2} i) = x^2 + x + 1$, and therefore that $A = B = C$.  (Note, however, that they are not necessarily equal to 1.)
