If $$f(x)$$ and $$g(x)$$ are two polynomials such that the polynomial $$P(x) = f(x^3) + xg(x^3)$$ is divisible by $$x^2 + x + 1$$, then $$P(1)$$ is equal to ___.

Since $$P(x) = f(x^3) + x\,g(x^3)$$ is divisible by $$x^2 + x + 1$$, both primitive cube roots of unity $$\omega$$ and $$\omega^2$$ (where $$\omega = e^{2\pi i/3}$$) are roots of $$P(x)$$. Note that $$\omega^3 = 1$$.

Substituting $$x = \omega$$: $$P(\omega) = f(\omega^3) + \omega\,g(\omega^3) = f(1) + \omega\,g(1) = 0$$. Substituting $$x = \omega^2$$: $$P(\omega^2) = f(1) + \omega^2\,g(1) = 0$$.

Subtracting these two equations gives $$(\omega - \omega^2)\,g(1) = 0$$. Since $$\omega \neq \omega^2$$, we conclude $$g(1) = 0$$. Substituting back into either equation gives $$f(1) = 0$$.

Therefore $$P(1) = f(1) + 1 \cdot g(1) = 0 + 0 = 0$$.