Question

Let $\omega$ be a solution of $x^3-1=0$ with $Im\left(\omega \right)>0$. lf $a=2$ 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. $-2$
• B. $2$
• C. $3$
• D. $-3$

Answer 1

The correct option is A $-2$

Since, $a,bandc$ satisfies
$\left[\begin{array}{ccc}a& b& c\end{array}\right]\left[\begin{array}{ccc}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$

So, we get the equations
$a+8b+7c=0$
$9a+2b+3c=0$
$7a+7b+7c=0$ or $a+b+c=0$
Since , $a=2$, so the equations become
$2+8b+7c=0$ .....(1)
$18+2b+3c=0$ .....(2)
$2+b+c=0$ .....(3)
Solving above equations, we get $b=12,c=-14$
$\frac{3}{{\omega }^a}+\frac{1}{{\omega }^b}+\frac{3}{{\omega }^c}$

$=\frac{3}{{\omega }^2}+\frac{1}{{\omega }^{12}}+\frac{3}{{\omega }^{-14}}$

$=\frac{3}{{\omega }^2}+\frac{1}{{\omega }^{12}}+3{\omega }^{14}$

$=\frac{3}{{\omega }^2}+\frac{1}{{\left({\omega }^3\right)}^4}+3{\left({\omega }^3\right)}^4{\omega }^2$

$=3\omega +1+3{\omega }^2$

$=3(\omega +{\omega }^2)+1$

Since $1+\omega +{\omega }^2=0$,

So, $\frac{3}{{\omega }^a}+\frac{1}{{\omega }^b}+\frac{3}{{\omega }^c}=3(-1)+1=-2$