If f(x) and g(x) are two polynomials such tha | vedclass

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(D) Given $P(x) = f(x^3) + xg(x^3)$.Since $P(x)$ is divisible by $x^2 + x + 1$,it must vanish at the roots of $x^2 + x + 1 = 0$. Let $\omega$ be a com

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(D) Given $P(x) = f(x^3) + xg(x^3)$.Since $P(x)$ is divisible by $x^2 + x + 1$,it must vanish at the roots of $x^2 + x + 1 = 0$. Let $\omega$ be a complex cube root of unity,then $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$.The roots are $\omega$ and $\omega^2$. Thus,$P(\omega) = 0$ and $P(\omega^2) = 0$.$P(\omega) = f(\omega^3) + \omega g(\omega^3) = f(1) + \omega g(1) = 0$  (Equation $1$)$P(\omega^2) = f((\omega^2)^3) + \omega^2 g((\omega^2)^3) = f(\omega^6) + \omega^2 g(\omega^6) = f(1) + \omega^2 g(1) = 0$  (Equation $2$)Subtracting Equation $2$ from Equation $1$:$(f(1) + \omega g(1)) - (f(1) + \omega^2 g(1)) = 0$$(\omega - \omega^2) g(1) = 0$Since $\omega 
eq \omega^2$,we must have $g(1) = 0$.Substituting $g(1) = 0$ into Equation $1$:$f(1) + \omega(0) = 0 \Rightarrow f(1) = 0$.We need to find $P(1)$:$P(1) = f(1^3) + 1 \cdot g(1^3) = f(1) + g(1) = 0 + 0 = 0$.

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 ....... .

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