Question

Consider a polynomial $p ( x )= x ^6+2 x ^2+1$. If $x _1, x _2, \ldots \ldots, x _6$ are the roots of the equation $p ( x )=0$ and $q ( x )= x ^3-1$, then find the value of $\displaystyle\prod_{i=1}^6 q \left( x _i\right)$

Answer 1

$\left(x-x_1\right)\left(x-x_2\right) \ldots \ldots\left(x-x_6\right)=x^6+2 x^2+1 $
$x =1 \left(1- x _1\right)\left(1- x _2\right) \ldots \ldots\left(1- x _6\right)=4 $ ...(1)
$x=w \left(w-x_1\right)\left(w-x_2\right) \ldots \ldots\left(w-x_6\right)=w^6+2 w^2+1=2\left(1+w^2\right)=-2 w$....(2)
$x = w ^2 \left( w ^2- x _1\right)\left( w ^2- x _2\right) \ldots \ldots\left( w ^2- x _6\right)= w ^{12}+2 w ^4+1=-2 w ^2 $.....(3)
$\text { Multiplying (1), (2) and (3) }$
$\left(1-x_1^3\right)\left(1-x_2^3\right) \ldots \ldots . .\left(1-x_6^3\right)=16 $
$ \left(1-x_1^3\right)\left(1-x_2^3\right) \ldots \ldots . .\left(1-x_6^3\right)=16 $
$\Rightarrow \left(x_1^3-1\right)\left(x_2^3-1\right) \ldots . .\left(x_6^3-1\right)=16 \Rightarrow \displaystyle\prod_{i=1}^6 q\left(x_i\right)=16 $