If f(x) and g(x) are two polynomials such

Solution:

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

$P(1)=f(1)+g(1) \ldots(1)$

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

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

$\mathrm{P}(\mathrm{w})=0=\mathrm{P}\left(\mathrm{w}^{2}\right)$ where $\mathrm{w}, \mathrm{w}^{2}$ are non-real cube

roots of units

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

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

$f(1)+w g(1)=2 \ldots(2)$

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

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

$(2)+(3)$

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

$2 f(1)=g(1) \ldots(4)$

$(2)-(3)$

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

$g(1)=0=f(1)$ from $(4)$

from (1) $P(1)=f(1)+g(1)=0$