If f ( x ) and g ( x ) are two polynomials such that the polynomial P ( x ) = f x 3 + x g x 3 is divisible…

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

$P\left(1\right)=f\left(1\right)+g\left(1\right) ...\left(1\right)$

Now $P\left(x\right)$ is divisible by $x^2+x+1$

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

$P\left(\omega \right)=0=P\left({\omega }^2\right)$ where $\omega ,{\omega }^2$ are non-real cube roots of units

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

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

$f\left(1\right)+\omega g\left(1\right)=0 ...\left(2\right)$

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

$f\left(1\right)+{\omega }^{2}g\left(1\right)=0 ...\left(3\right)$

Now, $\left(2\right)+\left(3\right)$

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

$2f\left(1\right)=g\left(1\right) ...\left(4\right)$

Now, $\left(2\right)-\left(3\right)$

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

$g\left(1\right)=0=f\left(1\right)$   from $\left(4\right)$

from $\left(1\right),$ $P\left(1\right)=f\left(1\right)+g\left(1\right)=0$