Let P(x) and Q(x) be two polynomials. Suppose that f(x) = P(x³) + xQ(x³) is divisible by x² + x + 1, then

We have,

${\mathrm{x}}^2+\mathrm{x}+1=(\mathrm{x}-\mathrm{\omega })\left(\mathrm{x}-{\mathrm{\omega }}^2\right)$

Since f(x) is divisible by x² + x + 1, $\mathrm{f}\left(\mathrm{\omega }\right)=0, \mathrm{f}\left({\mathrm{\omega }}^2\right)=0$, so

$\mathrm{P}\left({\mathrm{\omega }}^3\right)+\mathrm{\omega Q}\left({\mathrm{\omega }}^3\right)=0\Rightarrow \mathrm{P}\left(1\right)+\mathrm{\omega Q}\left(1\right)=0$             (1)

$\mathrm{P}\left({\mathrm{\omega }}^6\right)+{\mathrm{\omega }}^2\mathrm{Q}\left({\mathrm{\omega }}^6\right)=0\Rightarrow \mathrm{P}\left(1\right)+{\mathrm{\omega }}^2\mathrm{Q}\left(1\right)=0$          (2)

Solving Eqs. (1) and (2), we obtain

P(1) = 0 and Q(1) = 0

Therefore, both P(x) and Q(x) are divisible by x - 1. Hence, P(x³) and Q(x³) are divisible by x³ - 1 and so by x - 1 Since f(x) = P(x³) + xQ(x³), we get f(x) is divisible by x - 1.