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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
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Verified Answer
The correct answer is:
$p^3+q^3$
Given, $\quad x=p+q, \quad y+p \omega+q \omega^2 \quad$ and $z=p \omega^2+q \omega$
$\begin{aligned} \therefore x y z= & (p+q)\left(p \omega+q \omega^2\right)\left(p \omega^2+q \omega\right) \\ = & (p+q)\left(p^2 \omega^3+p q \omega^2+p q \omega^4+q^2 \omega^3\right) \\ = & (p+q)\left(p^2+p q \omega^2+p q \omega+q^2\right) \\ & \quad\left(\because \omega^3=1\right) \\ = & (p+q)\left(p^2+p q\left(\omega+\omega^2\right)+q^2\right) \\ = & (p+q)\left(p^2-p q+q^2\right)=p^3+q^3 \\ & \quad\left(\because 1+\omega+\omega^2=-1\right)\end{aligned}$
$\begin{aligned} \therefore x y z= & (p+q)\left(p \omega+q \omega^2\right)\left(p \omega^2+q \omega\right) \\ = & (p+q)\left(p^2 \omega^3+p q \omega^2+p q \omega^4+q^2 \omega^3\right) \\ = & (p+q)\left(p^2+p q \omega^2+p q \omega+q^2\right) \\ & \quad\left(\because \omega^3=1\right) \\ = & (p+q)\left(p^2+p q\left(\omega+\omega^2\right)+q^2\right) \\ = & (p+q)\left(p^2-p q+q^2\right)=p^3+q^3 \\ & \quad\left(\because 1+\omega+\omega^2=-1\right)\end{aligned}$
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