Question

If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $P ( x )=f\left( x ^{3}\right)+ xg \left( x ^{3}\right)$ is divisible by $x^{2}+x+1$, then $P(1)$ is equal to________.

Answer 1

$P ( x )=f\left( x ^{3}\right)+ xg \left( x ^{3}\right)$
$P (1)=f(1)+ g (1) \ldots$(1)
Now $P ( x )$ is divisible by $x ^{2}+ x +1$
$\Rightarrow P ( x )= Q ( x )\left( x ^{2}+ x +1\right)$
$P ( w )=0= P \left( w ^{2}\right)$ where $w , w ^{2}$ are non-real cube
roots of units
$P ( x )=f\left( x ^{3}\right)+ xg \left( x ^{3}\right)$
$P ( w )=f\left( w ^{3}\right)+ wg \left( w ^{3}\right)=0$
$f(1)+\operatorname{wg}(1)=2 \ldots(2)$
$P \left( w ^{2}\right)=f\left( w ^{6}\right)+ w ^{2} g \left( w ^{6}\right)=0$
$f(1)+ w ^{2} g (1)=0 \ldots(3)$
$(2)+(3)$
$\Rightarrow 2 f(1)+\left( w + w ^{2}\right) g (1)=0$
$2 f(1)= g (1) \ldots(4)$
$(2)-(3)$
$\Rightarrow \left( w - w ^{2}\right) g (1)=0$
$g(1)=0=f(1) $ from $(4)$
from (1) $P (1)=f(1)+ g (1)=0$