Question

If $x=a+b,y=a\omega +b{\omega }^2,z=a{\omega }^2+b\omega$, then xyz equals to where, $\omega$ is the cube root of unity.

• A. $a+b$
• B. $a^2+b^2$
• C. $a^3+b^3$
• D. $a^4+b^4$

Answer 1

The correct option is C $a^3+b^3$

Given : $x=a+b,y=a\omega +b{\omega }^2,z=a{\omega }^2+b\omega$

Consider, $xyz$

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

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

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

$=a^3+b^3+a^{2}b({\omega }^2+{\omega }^4+{\omega }^3)+a{b}^2({\omega }^3+{\omega }^2+{\omega }^4)$ ....... $[\because {\omega }^3=1]$

$=a^3+b^3+a^{2}b({\omega }^2+\omega +1)+a{b}^2({\omega }^2+\omega +1)$

$=a^3+b^3$ ..... $[\because ({\omega }^2+\omega +1)=0,{\omega }^3=1]$