Polynomials

As $h(x)$ is divisible by $x^2+x+1$ we can put $x=\omega$ and $x=\omega^2$ and it becomes

$h(\omega)= \omega f(\omega^3) + \omega^2 g(\omega^6)$ As $h(\omega)=0$

$\omega f(1) + \omega^2 g(1)=0$

Now $h(\omega^2)= \omega^2 f(\omega^6) + \omega^4 g(\omega^6)$ As $h(\omega^2)=0$

$\omega^2 f(1) + \omega g(1)=0$

now solving these two equations we get $f(1)=g(1)=0$

So the answer is zero.

Since $h(x)$ is divisible by $x^2+x+1$ , let $h(x) = q(x)(x^2+x+1)$ . Then we have:

$\begin{aligned} xf(x^3) + x^2g(x^6) & = q(x)(x^2+x+1) & \small \color{#3D99F6}{\text{Let }x = \omega \text{ the third complex root of unity}} \\ \omega f(\omega^3) + \omega^2g(\omega) & = q(\omega)(\omega^2+\omega+1) & \small \color{#3D99F6}{\text{Note that } \omega^2+\omega+1 = 0 \text{ and }\omega^3 = 1} \\ \omega f(1) + \omega^2g(1) & = q(\omega)(0) & \small \color{#3D99F6}{\text{Note that } \omega = \frac 12 + \frac {\sqrt 3} 2 i \text{ and }\omega^2 = \frac 12 - \frac {\sqrt 3} 2 i} \end{aligned}$

$\begin{aligned} \frac 12 \left(f(1)+g(1) \right) + \frac {\sqrt 3} 2 i \left(f(1)-g(1) \right) & = 0 \\ \implies f(1)+g(1) & = \boxed{0} \end{aligned}$