Question

Consider a polynomial $p(x)=x^6+2x^2+1$. If $x_1,x_2,\dots \dots ,x_6$ are the roots of the equation $p(x)=0$ and $q(x)=x^3-1$, then find the value of $\prod _{i=1}^{6}q\left(x_i\right)$

Answer 1

Answer: 16

$\left(x-x_1\right)\left(x-x_2\right)\dots \dots \left(x-x_6\right)=x^6+2x^2+1$
$x=1\left(1-x_1\right)\left(1-x_2\right)\dots \dots \left(1-x_6\right)=4$ ...(1)
$x=w\left(w-x_1\right)\left(w-x_2\right)\dots \dots \left(w-x_6\right)=w^6+2w^2+1=2\left(1+w^2\right)=-2w$....(2)
$x=w^2\left(w^2-x_1\right)\left(w^2-x_2\right)\dots \dots \left(w^2-x_6\right)=w^{12}+2w^4+1=-2w^2$.....(3)
$\text{ Multiplying (1), (2) and (3) }$
$\left(1-x_1^3\right)\left(1-x_2^3\right)\dots \dots ..\left(1-x_6^3\right)=16$
$\left(1-x_1^3\right)\left(1-x_2^3\right)\dots \dots ..\left(1-x_6^3\right)=16$
$\Rightarrow \left(x_1^3-1\right)\left(x_2^3-1\right)\dots ..\left(x_6^3-1\right)=16\Rightarrow \prod _{i=1}^{6}q\left(x_i\right)=16$