Question

Let $x=a+b,y=a\omega +b{\omega }^2,z=a{\omega }^2+b\omega$ where $\omega$ being cube root of unity. Then find the value of $xyz$

• A. $a^3-b^3$
• B. $a^3+b^3$
• C. $3ab$
• D. none of these

Answer 1

Answer: (B)

$a^3+b^3$

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

Now,

$xyz$

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

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

$=(a+b)\{a^2+ab(\omega +{\omega }^2)+b^2\}$ [ Since ${\omega }^3=1$]

$=(a+b)\{a^2-ab+b^2\}$ [ Since $1+\omega +{\omega }^2=0$]

$=a^3+b^3$.