Q86 / 160 · AP EAMCET 2014 · Mathematics
If $x=p+q, y=p \omega+q \omega^2 \quad$ and $\quad z=p \omega^2+q \omega$, where $\omega$ is a complex cube root of unity, then $x y z$ equals to
- Correct answerA$p^3+q^3$
- B$p^2-p q+q^3$
- C$1+p^3+q^3$
- D$p^3-q^3$
Solution
Given, $\quad x=p+q, \quad y+p \omega+q \omega^2 \quad$ and $z=p \omega^2+q \omega$
$\begin{aligned} \therefore x y z= & (p+q)\left(p \omega+q \omega^2\right)\left(p \omega^2+q \omega\right) \\ = & (p+q)\left(p^2 \omega^3+p q \omega^2+p q \omega^4+q^2 \omega^3\right) \\ = & (p+q)\left(p^2+p q \omega^2+p q \omega+q^2\right) \\ & \quad\left(\because \omega^3=1\right) \\ = & (p+q)\left(p^2+p q\left(\omega+\omega^2\right)+q^2\right) \\ = & (p+q)\left(p^2-p q+q^2\right)=p^3+q^3 \\ & \quad\left(\because 1+\omega+\omega^2=-1\right)\end{aligned}$
Hence, the correct answer is option: 'A'.