Let P(x) and Q(x) be two real polynomials. Suppose that $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{P}\left({\mathrm{x}}^3\right)+\mathrm{xQ}\left({\mathrm{x}}^3\right)$ is divisible by ${\mathrm{x}}^2+\mathrm{x}+1$, then

$\text{ Since }\mathrm{f}\left(\mathrm{x}\right)\text{ is divisible by }$${\mathrm{x}}^2+\mathrm{x}+1,\mathrm{f}\left(\mathrm{\omega }\right)=0,\mathrm{f}\left({\mathrm{\omega }}^2\right)=0$
$\therefore \mathrm{P}\left({\mathrm{\omega }}^3\right)+\mathrm{\omega Q}\left({\mathrm{\omega }}^3\right)=0\Rightarrow \mathrm{P}\left(1\right)+\mathrm{\omega Q}\left(1\right)=0$
and $\mathrm{P}\left({\mathrm{\omega }}^6\right)+{\mathrm{\omega }}^2\mathrm{Q}\left({\mathrm{\omega }}^6\right)=0\Rightarrow \mathrm{P}\left(1\right)+{\mathrm{\omega }}^2\mathrm{Q}\left(1\right)=0\dots \left(1\right)$
$\text{ and }\mathrm{P}\left({\mathrm{\omega }}^6\right)+{\mathrm{\omega }}^2\mathrm{Q}\left({\mathrm{\omega }}^6\right)=0\Rightarrow \mathrm{P}\left(1\right)+{\mathrm{\omega }}^2\mathrm{Q}\left(1\right)=0\dots \left(2\right)$
Solving (1) and (2) we obtain $\mathrm{P}\left(1\right)=0\text{ and }\mathrm{Q}\left(1\right)=0\text{. }$
$\therefore \text{ Both }\mathrm{P}\left(\mathrm{x}\right)\text{ and }\mathrm{Q}\left(\mathrm{x}\right)\text{ are divisible by }\mathrm{x}-1$.
$\Rightarrow \mathrm{P}\left({\mathrm{x}}^3\right)\text{ and }\mathrm{Q}\left({\mathrm{x}}^3\right)\text{ are divisible by }{\mathrm{x}}^3-1$ and hence by x – 1.
Since $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{P}\left({\mathrm{x}}^3\right)+\mathrm{xQ}\left({\mathrm{x}}^3\right),\text{ we get }\mathrm{f}\left(\mathrm{x}\right)\text{ is divisible by }$by x – 1.
Since f(x) = P(x³) + xQ(x³), we get f(x) is divisible by x – 1.