1. Let $P(x)=f(x^3)+xg(x^3).$ We are given that $P(x)$ is divisible by $x^2+x+1.$

2. The roots of $x^2+x+1=0$ are the non-real cube roots of unity: $\\omega,\\omega^2,$ where $\\omega^3=1,\quad \\omega\neq 1,\quad 1+\\omega+\\omega^2=0.$

Since $x^2+x+1$ divides $P(x)$, we must have $P(\\omega)=0 \quad \text{and} \quad P(\\omega^2)=0.$

3. Compute $P(\\omega)$: $P(\\omega)=f(\\omega^3)+\\omega g(\\omega^3)=f(1)+\\omega g(1).$ Thus, f(1)+\\omega g(1)=0. \tag{1}

Similarly, P(\\omega^2)=f((\\omega^2)^3)+\\omega^2 g((\\omega^2)^3)=f(1)+\\omega^2 g(1)=0. \tag{2}

4. Subtract (1) and (2): $(\\omega-\\omega^2)g(1)=0.$ Since $\\omega\neq \\omega^2$, we get $g(1)=0.$

Substitute into (1): $f(1)=0.$

5. Now compute $P(1)$: $P(1)=f(1^3)+1\cdot g(1^3)=f(1)+g(1)=0+0=0.$

Therefore, $\boxed{P(1)=0}.$