The polynomial ${\mathrm{x}}^6+4{\mathrm{x}}^5+3{\mathrm{x}}^4+2{\mathrm{x}}^3+\mathrm{x}+1$ is divisible by (where $\omega$ is the cube root of unity)

where $\omega$ is one of the imaginary cube roots of unity.

Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^6+4{\mathrm{x}}^5+3{\mathrm{x}}^4+2{\mathrm{x}}^3+\mathrm{x}+1$. Hence,

$\begin{array}{c}\mathrm{f}\left(\mathrm{\omega }\right)={\mathrm{\omega }}^6+4{\mathrm{\omega }}^5+3{\mathrm{\omega }}^4+2{\mathrm{\omega }}^3+\mathrm{\omega }+1\\ =1+4{\mathrm{\omega }}^2+3\mathrm{\omega }+2+\mathrm{\omega }+1\\ =4\left({\mathrm{\omega }}^2+\mathrm{\omega }+1\right)\\ =0\end{array}$

Hence, f(x) is divisible by x - $\omega$. Then f(x) is also divisible by x - $\omega$² (as complex roots occur in conjugate pairs).

$\begin{array}{l}\mathrm{f}(-\mathrm{\omega })=(-\mathrm{\omega }{)}^6+4(-\mathrm{\omega }{)}^5+3(-\mathrm{\omega }{)}^4+2(-\mathrm{\omega }{)}^3+(-\mathrm{\omega })+1\\ ={\mathrm{\omega }}^6-4{\mathrm{\omega }}^5+3{\mathrm{\omega }}^4-2{\mathrm{\omega }}^3-\mathrm{\omega }+1\\ =1-4{\mathrm{\omega }}^2+3\mathrm{\omega }-2-\mathrm{\omega }+1\\ \ne 0\end{array}$