Complex Numbers And Quadratic Equations question 249
Question: If $ x=a+b,y=a\omega +b{{\omega }^{2}},z=a{{\omega }^{2}}+b\omega $ , then the value of $ x^{3}+y^{3}+z^{3} $ is equal to [Roorkee 1977; IIT 1970]
Options:
A) $ a^{3}+b^{3} $
B) $ 3(a^{3}+b^{3}) $
C) $ 3(a^{2}+b^{2}) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ x^{3}+y^{3}+z^{3}={{(a+b)}^{3}}+{{(a\omega +b{{\omega }^{2}})}^{3}}+{{(a{{\omega }^{2}}+b\omega )}^{3}} $ $ =3a^{3}+3b^{3}+3(a^{2}b+ab^{2})(1+{{\omega }^{2}}{{\omega }^{2}}+\omega {{\omega }^{4}}) $ $ =3a^{3}+3b^{3}+3(a^{2}b+ab^{2})(1+\omega +{{\omega }^{2}}) $ $ = $ $ 3(a^{3}+b^{3}) $ Trick: As in the previous question $ x^{3}+y^{3}+z^{3}={{(4)}^{3}}+{{(-2)}^{3}}+{{(-2)}^{3}}=48 $ and (b) i.e. $ 3(a^{3}+b^{3})=48 $
BETA
