Question

Let $\omega$ be a solution of $x^3-1=0$ with Im($\omega$) > 0. If $a=2$ with b and c satisfying (E), then the value of $\frac{3}{{\omega }^a}+\frac{1}{{\omega }^b}+\frac{3}{{\omega }^c}$ is equal to

• A. -1
• B. -2
• C. 3
• D. -3

Answer 1

The correct option is B -2

$\left[\begin{array}{ccc}a& b& c\end{array}\right]\left[\begin{array}{ccc}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$
$\Rightarrow a+8b+7c=0$ .........(i)
and $9a+2b+3c=0$ ..........(ii)
and $7a+7b+7c=0$ .........(iii)
$\frac{a}{1}=\frac{b}{6}=\frac{c}{7}=k\left(say\right)$
$\Rightarrow a=k,b=6k,c=-7k$
Given a=2
$\Rightarrow k=2$
$\Rightarrow b=12,c=-14$
Now, $\frac{3}{{\omega }^a}+\frac{1}{{\omega }^b}+\frac{3}{{\omega }^c}$
$=\frac{3}{{\omega }^2}+\frac{1}{{\omega }^{12}}+\frac{3}{{\omega }^{-14}}$
$=\frac{3\omega }{{\omega }^3}+\frac{1}{({\omega }^3{)}^4}+3{\omega }^{14}$
$=-3+1=-2$