If f(x) and g(x) are two polynomials such that the polynomial

$\mathrm{P}(\mathrm{x})=f\left(\mathrm{x}^{3}\right)+\mathrm{xg}\left(\mathrm{x}^{3}\right)$

$\mathrm{P}(1)=f(1)+\mathrm{g}(1)$            ......(1)

Now $\mathrm{P}(\mathrm{x})$ is divisible by $\mathrm{x}^{2}+\mathrm{x}+1$

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

roots of units

$\mathrm{P}(\mathrm{x})=f\left(\mathrm{x}^{3}\right)+\mathrm{xg}\left(\mathrm{x}^{3}\right)$

$\mathrm{P}(\mathrm{w})=f\left(\mathrm{w}^{3}\right)+\mathrm{wg}\left(\mathrm{w}^{3}\right)=0$

$f(1)+w g(1)=2$         ......(2)

$\mathrm{P}\left(\mathrm{w}^{2}\right)=f\left(\mathrm{w}^{6}\right)+\mathrm{w}^{2} \mathrm{~g}\left(\mathrm{w}^{6}\right)=0$

$f(1)+w^{2} g(1)=0$                 ......(3)

$(2)+(3)$

$\Rightarrow 2 f(1)+\left(w+w^{2}\right) g(1)=0$

$2 f(1)=g(1)$     .....(4)

$(2)-(3)$

$\Rightarrow\left(w-w^{2}\right) g(1)=0$

$\mathrm{g}(1)=0=f(1)$ from (4)

from (1) $\mathrm{P}(1)=f(1)+\mathrm{g}(1)=0$