logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  TN XII Math  >>  Complex Numbers
0 votes

Prove that if $\omega^3=1$,then (i) $(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=a^3+b^3+c^3-3abc$

This is the first part of the multi-part question Q3

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If $\omega$ is a cube root of unity then $\omega^3=1$ and $1+\omega+\omega^2=0$
  • $\omega^4=\omega$
  • $\omega+\omega^2=-1$
$(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=(a+b+c)(a^2+ab\omega^2+ac\omega+ab\omega+b^2\omega^3+bc\omega^2+ca\omega^2+bc\omega^4+c^2\omega^3)$
$\Rightarrow (a+b+c)(a^2+ab(\omega+\omega^2)+ac(\omega+\omega^2)+bc(\omega+\omega^2)+b^2+c^2)$
We know that $\omega+\omega^2=-1$
$\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
We know that $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$
$\Rightarrow a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Therefore $(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=a^3+b^3+c^3-3abc$
Hence proved.
answered Jun 12, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...