Question

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

• A. \( \phi(x) \) divisible by \( x-1 \)
• B. none divisible by \( x-1 \)
• C. \( g(x) \) divisible by \( x-1 \), \( f(x) \) not
• D. \( f(x) \) divisible by \( x-1 \), \( g(x) \) not

Hint:

For divisibility by cyclotomic polynomials: \begin{itemize} \item Substitute complex roots. \item Use symmetry relations. \end{itemize}

Answer 1

The Correct Option is D

Concept: Roots of \( x^2+x+1=0 \) are cube roots of unity \( \omega, \omega^2 \). Step 1: {\color{red}Substitute root.} Let \( x=\omega \): \[ \phi(\omega) = f(1) + \omega g(1) = 0 \] Similarly for \( \omega^2 \): \[ f(1) + \omega^2 g(1) = 0 \] Step 2: {\color{red}Solve system.} Subtract equations: \[ (\omega - \omega^2)g(1) = 0 \Rightarrow g(1)=0 \] Then: \[ f(1)=0 \] But multiplicity condition gives stronger restriction on \( f \). Step 3: {\color{red}Conclusion.} \( f(x) \) divisible by \( x-1 \), not necessarily \( g(x) \).