<s>[INST] 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?
 [/INST] Yes, I just check your calculus and are all correct. Your polynomial of grade $3$ defined as $x^3+3x^2+3x+2=0$ has one real root $x_{1}=-2$ and two complex roots $x_{2,3}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ so it can be written as $P(x)=(x+2)(x-(\frac{1}{2}-\frac{\sqrt{3}}{2}i))(x-(\frac{1}{2}+\frac{\sqrt{3}}{2}i))$. Then, remember the following: "a property of the second degree equations with real coefficients that have complex roots is that they are conjugated to each other". So, knowing this, there is no way you will have in common with the other polynomial one real root and other complex (is the only other option we have available), then: $a=b=c=1$.
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