Complex Numbers And Quadratic Equations question 261
Question: If $ 1,\omega ,{{\omega }^{2}} $ are the three cube roots of unity, then $ {{(3+{{\omega }^{2}}+{{\omega }^{4}})}^{6}}= $ [MP PET 1995]
Options:
A) 64
B) 729
C) 2
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{(3+{{\omega }^{2}}+{{\omega }^{4}})}^{6}}={{(3+{{\omega }^{2}}+\omega )}^{6}}={{(3-1)}^{6}}=64 $
BETA
