If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x³) + x g(x³) is divisible by x² + x + 1, then P(1) is equal to ___________.

Given, p(x) = f(x³) + xg(x³)<br><br>We know, x² + x + 1 = (x $-$ $\omega$) (x $-$ $\omega$²)<br><br>Given, p(x) is divisible by x² + x + 1. So, roots of p(x) is $\omega$ and $\omega$².<br><br>As root satisfy the equation,<br><br>So, put x = $\omega$<br><br>p($\omega$) = f($\omega$³) + $\omega$g($\omega$³) = 0<br><br>= f(1) + $\omega$g(1) = 0 [$\omega$³ = 1]<br><br>= f(1) + $\left( { - {1 \over 2} + {{i\sqrt 3 } \over 2}} \right)$ g(1) = 0<br><br>$\Rightarrow$ f(1) $-$ ${{g(1)} \over 2} + i\left( {{{\sqrt 3 g(1)} \over 2}} \right)$ = 0 + i0<br><br>Comparing both sides, we get<br><br>f(1) $-$ ${{g(1)} \over 2}$ = 0<br><br>and ${{{\sqrt 3 } \over 2}g(1) = 0}$ $\Rightarrow$ g(1) = 0<br><br>So, f(1) = 0<br><br>Now, p(1) = f(1) + 1 . g(1) = 0 + 0 = 0