Question

What is \[{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}}\] equal to, where $\omega $ is the cube root of unity?
A) $1$
B) $3\omega $
C) $3{\omega ^2}$
D) $0$

Answer 1

Hint: To solve this question, we will first use the condition that $\omega $ is the cube root of unity. Using this we will make all possible equations which we can take so that we can solve the given expression. Now, after arranging the given expression into required form, we will get our answer out of four multiple choices given to us.

Complete step-by-step answer:
We have been given that $\omega $ is the cube root of unity. We need to find the value of \[{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}}\].
So, it is given that, $\omega $ is the cube root of unity.
i.e., $\omega = {(1)^{\dfrac{1}{3}}}$
On taking cubes on both sides, we get
$
\Rightarrow {(\omega )^3} = {({(1)^{\dfrac{1}{3}}})^3} \\
\Rightarrow {(\omega )^3} = 1......eq.(1) \\
\Rightarrow {(\omega )^3} - 1 = 0 \\
\Rightarrow {(\omega )^3} - {(1)^3} = 0 \\
\Rightarrow (\omega - 1)[{(\omega )^2} + \omega + 1)] = 0 \\
\Rightarrow (\omega - 1) = 0,[{(\omega )^2} + \omega + 1)] = 0.....eq.(2) \\
$
Now, let us find the value of \[{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}}\]. Firstly, we will modify the given expression \[{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}}\] in the form of above solved equations.
\[ = {({\omega ^3})^{33}}\omega + {({\omega ^3})^{66}}{\omega ^2} + {({\omega ^3})^{100}}\]
Now, using \[eq.{\text{ }}\left( 1 \right),\] we get
\[ = {(1)^{33}}\omega + {(1)^{66}}{\omega ^2} + {(1)^{100}}\]
\[ = {(1)^{33}}\omega + {(1)^{66}}{\omega ^2} + {(1)^{100}}\]
\[ = \omega + {\omega ^2} + 1\]
Now, using \[eq.\left( 2 \right),\] we get
\[ = 0\]
So, \[{\omega ^{100}} + {\omega ^{200}} + {\omega ^{300}} = 0\]

Note: Students might get confused with the statement, i.e., $\omega $ is the cube root of unity. So, you can do one thing, just go with the words. i.e., by cube root of unity, they mean power $\dfrac{1}{3}$ of \[1.\] On the left side take omega, and on right side take cube root of unity. Now, we took a cube on both sides, so we got one cube of omega, and on the other side we got a cube of cube root of unity, which cancels the root, and we got $1$ on the other side. Then, with that we solved further.