Complex Numbers And Quadratic Equations question 267
Question: If $ 1,\omega ,{{\omega }^{2}} $ are three cube roots of unity, then $ {{(a+b\omega +c{{\omega }^{2}})}^{3}} $ + $ {{(a+b{{\omega }^{2}}+c\omega )}^{3}} $ is equal to, if $ a+b+c=0 $ [West Bengal JEE 1992]
Options:
A) $ 27abc $
B) 0
C) $ 3abc $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Trick: Put $ a=1,b=1,c=-2 $ , $ \because $ $ a+b+c=0 $
$ \therefore {{(1+\omega -2{{\omega }^{2}})}^{3}}+{{(1+{{\omega }^{2}}-2\omega )}^{3}} $ $ ={{(-3{{\omega }^{2}})}^{3}}+{{(-3\omega )}^{3}}=-27-27=-54 $ Also option (a) gives the value $ i.e., $ $ 27\times 1\times 1(-2)=-54 $
BETA
