Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

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

1 Answer

Comment
A)
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.
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...