Question

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

• A. $10$
• B. $16$
• C. $18$
• D. $20$

Answer 1

The correct option is B $16$

Given : $p\left(x\right)=x^6+2x^2+1$
$\Rightarrow p\left(x\right)=(x-x_1)(x-x_2)\cdots (x-x_6)$
Also,
$q\left(x\right)=x^3-1$
This can be written as
$q\left(x\right)=(x-1)(x-\omega )(x-{\omega }^2)$
Now,
$p\left(1\right)=(1-x_1)(1-x_2)\cdots (1-x_6)\Rightarrow (x_1-1)(x_2-1)\cdots (x_6-1)=4p\left(\omega \right)=(\omega -x_1)(\omega -x_2)\cdots (\omega -x_6)\Rightarrow (x_1-\omega )(x_2-\omega )\cdots (x_6-\omega )={\omega }^6+2{\omega }^2+1\Rightarrow (x_1-\omega )(x_2-\omega )\cdots (x_6-\omega )=2(1+{\omega }^2)\Rightarrow (x_1-\omega )(x_2-\omega )\cdots (x_6-\omega )=-2\omega p\left({\omega }^2\right)=(x_1-{\omega }^2)(x_2-{\omega }^2)\cdots (x_6-{\omega }^2)\Rightarrow (x_1-{\omega }^2)(x_2-{\omega }^2)\cdots (x_6-{\omega }^2)={\omega }^{12}+2{\omega }^4+1\Rightarrow (x_1-{\omega }^2)(x_2-{\omega }^2)\cdots (x_6-{\omega }^2)=2(1+\omega )\Rightarrow (x_1-{\omega }^2)(x_2-{\omega }^2)\cdots (x_6-{\omega }^2)=-2{\omega }^2$

Therefore,
$\prod _{i=1}^{6}q\left(x_i\right)=4\times (-2{\omega }^2)\times (-2\omega )\Rightarrow \prod _{i=1}^{6}q\left(x_i\right)=16$


Alternate solution:
$p\left(x\right)=x^6+2x^2+1$
$x_i$ is a root of $p\left(x\right)$ for $i=1,2,\dots ,6$
Let $Q\left(x\right)=0$ has roots as $x_i^3-1$, so
$Q\left(x\right)=(\sqrt[3]{3x+1}{)}^6+2(\sqrt[3]{3x+1}{)}^2+1\Rightarrow (\sqrt[3]{3x+1}{)}^6+2(\sqrt[3]{3x+1}{)}^2+1=0\Rightarrow x^2+2x+1+1=-2(x+1{)}^{2/3}\Rightarrow (x^2+2x+2{)}^3+8(x+1{)}^2=0\prod _{i=1}^{6}q\left(x_i\right)=\text{product of roots of the above equation}\Rightarrow \prod _{i=1}^{6}q\left(x_i\right)=2^3+8(1{)}^2=16$