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

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

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

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

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

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

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

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

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

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

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

Now, $( 2 ) + ( 3 )$

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

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

Now, $( 2 ) - ( 3 )$

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

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

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