Question

If $1,\omega ,{\omega }^2$ are the cube roots of unity, prove that:
$(a+b)(a\omega +b{\omega }^2)(a{\omega }^2+b\omega )=a^3+b^3$

Answer 1

Given:- $1,\omega ,{\omega }^2$ are the cube roots of unity

To prove:- $(a+b)(a\omega +b{\omega }^2)(a{\omega }^2+b\omega )=a^3+b^3$

Proof:-

Since $1,\omega$ and ${\omega }^2$ are the cube roots of unity.

Therefore,

$1+\omega +{\omega }^2=0$

${\omega }^3=1$

${\omega }^{3n+1}=\omega$

Now,

$(a+b)(a\omega +b{\omega }^2)(a{\omega }^2+b\omega )$

$=(a^2\omega +b^2{\omega }^2+ab(\omega +{\omega }^2))(a{\omega }^2+b\omega )$

$=(a^2\omega +b^2{\omega }^2-ab)(a{\omega }^2+b\omega )$

$=a^3{\omega }^3+a^{2}b{\omega }^2+b^{2}a{\omega }^4+b^3{\omega }^3-a^{2}b{\omega }^2-a{b}^2\omega$

$=a^3-a^{2}b{\omega }^2+a{b}^2\omega +b^3-a^{2}b{\omega }^2-a{b}^2\omega$

$=a^3+b^3$

Hence proved.