$Q(x)=f(x^3)+xg(x^3)$ and $Q(x)$ is divisible by $x^2+x+1$ Let $Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$ and $Q(x)$ is divisible by $x^2+x+1$, then find the value of $Q(1)$
My approach is as follow
$Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$
$Q\left( x \right) = \phi \left( x \right)\left( {{x^2} + x + 1} \right)$
$Q\left( \omega  \right) = \phi \left( \omega  \right)\left( {{\omega ^2} + \omega  + 1} \right) = 0$
$Q\left( {{\omega ^2}} \right) = \phi \left( \omega  \right)\left( {{\omega ^4} + {\omega ^2} + 1} \right) = 0$
$Q\left( \omega  \right) = f\left( {{\omega ^3}} \right) + \omega g\left( {{\omega ^3}} \right) = f\left( 1 \right) + \omega g\left( 1 \right) = 0$
$Q\left( {{\omega ^2}} \right) = f\left( {{\omega ^6}} \right) + {\omega ^2}g\left( {{\omega ^6}} \right) = f\left( 1 \right) + {\omega ^2}g\left( 1 \right) = 0$
Adding we get $2f(1)-g(1)=0$
$Q\left( 1 \right) = f\left( 1 \right) + g\left( 1 \right)$
Not able to proceed from here

Hint: if $x^2+x+1|Q(x)$, then
$$x^3-1\ \big| (x-1)Q(x)=(x^2-x)g(x^3)+(x-1)f(x^3).$$
See if you can use this, combined with the property that
$$f(x)-g(x)\ \big|\ h(f(x))-h(g(x))$$
for any polynomials $f,g,h$, to find $f(1)$ and $g(1)$.
