Question

$\omega$ is a cube root of unity, $\omega \ne 1$.

Match the following.

(1) ${\omega }^{99}$ (p) - 1

(2) 1 + ${\omega }^{33}$ + ${\omega }^{66}$ (q) 0

(3) 1 + ${\omega }^{14}$ + ${\omega }^{28}$ (r) 3

(4) ${\omega }^4$ + ${\omega }^5$ + ${\omega }^6$ (s) 1

• A. 1 - p, 2 - q, 3 - R, 4 - s
• B. 1 - s, 2 - q, 3 - r, 4 - q
• C. 1 - s, 2 - r, 3 - q, 4 - q
• D. 1 - p, 2 - q, 3 - q, 4 - q

Answer 1

The correct option is C

1 - s, 2 - r, 3 - q, 4 - q

Let's look at the following relations
If $\omega$ is the cube root of unity , then ${\omega }^3$ = 1
We can write the same as $\omega$ $\times$ ${\omega }^2$ = 1
$\Rightarrow$ $\omega$ = $\frac{1}{{\omega }^2}$
1, $\omega$ and ${\omega }^2$ are the roots of the equation $x^3$ -1 = 0
Sum of the roots = 1 + $\omega$ + ${\omega }^2$ = -$\frac{\left(coefficientof{x}^2\right)}{coefficientof{x}^3}$
= $\frac{0}{1}$=0
$\Rightarrow$ 1 + $\omega$ + ${\omega }^2$ = 0
We will solve this question using these two relations
(1) ${\omega }^{99}$ = ${\left({\omega }^3\right)}^{33}$ = 1
(2) 1 + ${\omega }^{33}$ + ${\omega }^{66}$ = 1 + ${\left({\omega }^3\right)}^{11}$ + ${\left({\omega }^3\right)}^{22}$ = 3
(3) 1 + ${\omega }^{14}$ + ${\omega }^{28}$ = 1 + ${\omega }^{12}$${\omega }^2$ + ${\omega }^{27}\omega$
= 1 + ${\omega }^2$ + $\omega$
= 0
(4) ${\omega }^4$ + ${\omega }^5$ + ${\omega }^6$ = ${\omega }^4$( 1+ $\omega$ + ${\omega }^2$) = 0