Question

If $\omega$ is a cube root of unity, and the value of
$(1-\omega )(1-{\omega }^2)(1-{\omega }^4)(1-{\omega }^8)=\alpha ,$ and
$\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}=\beta$ then

• A. $\alpha +\beta =8$
• B. $\alpha \beta =9$
• C. $\alpha \beta =-9$
• D. $\alpha +\beta =-9$

Answer 1

Answer: (A), (C)

$\alpha \beta =-9$

$(1-\omega )(1-{\omega }^2)(1-{\omega }^4)(1-{\omega }^8)$
$=(1-\omega )(1-{\omega }^2)(1-\omega )(1-{\omega }^2)[\because {\omega }^3=1]$
$=(1-{\omega }^2{)}^2(1-\omega {)}^2$
$=(1+{\omega }^4-2w^2)(1+{\omega }^2-2\omega )[\because {\omega }^3=1]$
$=(1+\omega -2{\omega }^2)(-\omega -2\omega )[\because 1+\omega +{\omega }^2=1]$
$=(-3{\omega }^2)(-3\omega )$
$=9=\alpha \cdots \left(1\right)$

Now,
$\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}$
$=\frac{a{\omega }^3+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{a+b\omega +c{\omega }^2}{c{\omega }^3+a\omega +b{\omega }^2}[\because {\omega }^3=1]=\omega +{\omega }^2[\because 1+\omega +{\omega }^2=1]=-1=\beta \cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\alpha +\beta =8$
and, $\alpha \times \beta =-9$