Given: \[ \log (1 - x + x^2) = a_1x + a_2x^2 + a_3x^3 + \dots + a_nx^n \] Substitute \(x = 1, \, \omega, \, \omega^2\) respectively:\[\] 1. For \(x = 1\): \[ \log(1) = a_1 + a_2 + a_3 + \dots + a_n \] 2. For \(x = \omega\): \[ \log(1 - \omega + \omega^2) = a_1\omega + a_2\omega^2 + a_3\omega^3 + \dots + a_n\omega^n \] 3. For \(x = \omega^2\): \[ \log(1 - \omega^2 + \omega^4) = a_1\omega^2 + a_2\omega^4 + a_3\omega^6 + \dots + a_n\omega^{2n} \] Adding the above three equations gives: \[ \log(-2\omega) + \log(-2\omega^2) = a_1(1 + \omega + \omega^2) + a_2(1 + \omega^2 + \omega^4) + \dots + a_n(1 + \omega^n + \omega^{2n}) \] Since \(1 + \omega + \omega^2 = 0\), all coefficients except those involving \(a_3, a_6, a_9, \dots\) will vanish. Thus: \[ \log(-2\omega \times -2\omega^2) = 3(a_3 + a_6 + a_9 + \dots) \] Simplify: \[ \log(4) = 3(a_3 + a_6 + a_9 + \dots) \] So: \[ a_3 + a_6 + a_9 + \dots = \frac{2}{3} \log 2 \]