Complex Numbers And Quadratic Equations question 666
Question: The value of $ {{(1+2\omega +{{\omega }^{2}})}^{3n}}-{{(1+\omega +2{{\omega }^{2}})}^{3n}} $ is:
Options:
A) 0
B) 1
C) $ \omega $
D) $ {{\omega }^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
We have, $ {{(1+{{\omega }^{2}}+2\omega )}^{3n}}-{{(1+\omega +2{{\omega }^{2}})}^{3n}} $ We know that, $ 1+\omega +{{\omega }^{2}}=0and{{\omega }^{3}}=1 $
$ \therefore $ given expression is equal to $ {{(2\omega -\omega )}^{3n}}-{{(2{{\omega }^{2}}-{{\omega }^{2}})}^{3n}} $ $ = {{(\omega )}^{3n}}- {{( {{\omega }^{2}} )}^{3n}} = {{( {{\omega }^{3}} )}^{n}} - {{( {{\omega }^{3}} )}^{2n}} =1-1=0 $
BETA
