# 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

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.