Complex Numbers And Quadratic Equations question 248
Question: If $ x=a+b,y=a\alpha +b\beta $ and $ z=a\beta +b\alpha , $ where $ \alpha $ and $ \beta $ are complex cube roots of unity, then $ xyz $ = [IIT 1978; Roorkee 1989; RPET 1997]
Options:
A) $ a^{2}+b^{2} $
B) $ a^{3}+b^{3} $
C) $ a^{3}b^{3} $
D) $ a^{3}-b^{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
If $ x=a+b,y=a\alpha +b\beta $ and $ z=\alpha \beta +b\alpha $ Then $ xyz=(a+b)(a\omega +b{{\omega }^{2}})(a{{\omega }^{2}}+b\omega ), $ where $ \alpha =\omega $ and $ \beta ={{\omega }^{2}} $ $ =(a+b)(a^{2}+ab{{\omega }^{2}}+ab\omega +b^{2}) $ $ =(a+b)(a^{2}-ab+b^{2})=a^{3}+b^{3} $ Trick: Put $ a=b=2 $ then $ x=4,y=2(\omega +{{\omega }^{2}})=-2 $ and $ z=2({{\omega }^{2}}+\omega )=-2 $
$ \therefore $ $ xyz=4(-2)(-2)=16 $ and (b) i.e. $ a^{3}+b^{3}=16 $ .
BETA
