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(A) Given that $\phi(x) = f(x^3) + x g(x^3)$ is divisible by $x^2 + x + 1$. Let $\omega$ be a complex cube root of unity, then $\omega^2 + \omega + 1

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$\phi(x)$ is divisible by $(x-1)$

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(A) Given that $\phi(x) = f(x^3) + x g(x^3)$ is divisible by $x^2 + x + 1$. Let $\omega$ be a complex cube root of unity, then $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$. Since $\phi(x)$ is divisible by $x^2 + x + 1$, we have $\phi(\omega) = 0$. $\phi(\omega) = f(\omega^3) + \omega g(\omega^3) = f(1) + \omega g(1) = 0$. Similarly, $\phi(\omega^2) = f(\omega^6) + \omega^2 g(\omega^6) = f(1) + \omega^2 g(1) = 0$. Subtracting the two equations: $(\omega - \omega^2) g(1) = 0$. Since $\omega 
eq \omega^2$, we must have $g(1) = 0$. Substituting $g(1) = 0$ into $f(1) + \omega g(1) = 0$, we get $f(1) = 0$. Since $f(1) = 0$ and $g(1) = 0$, both $f(x)$ and $g(x)$ are divisible by $(x-1)$.

If $f(x)$ and $g(x)$ are two polynomials such that $\phi(x) = f(x^3) + x g(x^3)$ is divisible by $x^2 + x + 1$, then

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