Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$ Find the common roots of $x^3 + 2x^2 + 2x + 1$ and $x^{1990} + x^{200} + 1$.
I completed the first part of the question by finding the roots of the first equation. I obtained $3$ roots, one of them being $-1$ and the other two complex. It is evident that $-1$ is not a root of the second equation, but how can I find out whether the other two roots are common or not?
 A: The complex roots of the first equation satisfy $x^2+x+1=0$.
What are the roots of this equation, and how do their powers behave?
A: The roots of $x^2+x+1$ are $\omega$ and $\omega^2$ (The two complex cube roots of unity)
and $\omega^3=1$. Therefore
$$ \omega^{1990}+\omega^{200}+1=\omega+\omega^2+1=0 $$
and
$$ \omega^{2\times1990}+\omega^{2\times200}+1=\omega^2+\omega+1=0$$
So $\omega$ and $\omega^2$ are the common roots.
A: $$x^2+x+1=0$$
Let's just consider$$x=\exp\left(\frac{i\pi}{3}\right).$$
$$1990 \equiv 1989+1 \equiv 1 \mod 3$$
$$200 \equiv 198+2 \equiv 2 \mod 3$$
Hence $$x^{1990}+x^{200}+1 = x+x^2+1=0$$
$\exp\left(-\frac{i\pi}{3} \right)$ being the conjugate must be another solution as well.
A: The solution to the equation $x^2+x+1=0$  is  $e^{\frac{2\pi}{3}}$ and $e^{\frac{4\pi}{3}}$ .Having $1990=3\times 663 +1$ and $200=3\times 66 +2 $, we can evaluate the second equation as 
if $ x= e^{\frac{2\pi}{3}} $ then $x^{1990}=x^{3\times 663 +1}=x,x^{200}=x^{3\times 66 +2}=x^2$, so $e^{\frac{2\pi}{3}}$ is root of second equation.
by the property of real polynomials, $\textit{conjugate of a root is also a root of that polynomial}$ . 
So, the two complex solution of the first equation is also solution for the second one.
A: $$x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1=(x+1)(x^2+x+1).$$
We see that $-1$ it's not common root and
$$x^{1990}+x^{200}+1=x^{1990}-x+x^{200}-x^2+x^2+x+1=$$
$$=x((x^3)^{663}-1)+x^2((x^3)^{66}-1)+x^2+x+1\equiv0(\mod(x^2+x+1)),$$
which says that roots of $x^2+x+1$ are all common roots and we get the answer:
$$\left\{-\frac{1}{2}+\frac{\sqrt3}{2}i,-\frac{1}{2}-\frac{\sqrt3}{2}i\right\}$$
