If x=p+q,y=p omega+q omega^2 and z=p omega^2+ | vedclass

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(A) Given,$x=p+q$,$y=p \omega+q \omega^2$ and $z=p \omega^2+q \omega$.We know that $\omega^3=1$ and $1+\omega+\omega^2=0$,which implies $\omega+\omega

Vedclass · Accepted Answer

$p^3+q^3$

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(A) Given,$x=p+q$,$y=p \omega+q \omega^2$ and $z=p \omega^2+q \omega$.We know that $\omega^3=1$ and $1+\omega+\omega^2=0$,which implies $\omega+\omega^2=-1$.Now,$xyz = (p+q)(p \omega+q \omega^2)(p \omega^2+q \omega)$$xyz = (p+q)(p^2 \omega^3 + pq \omega^2 + pq \omega^4 + q^2 \omega^3)$$xyz = (p+q)(p^2(1) + pq \omega^2 + pq \omega + q^2(1))$$xyz = (p+q)(p^2 + pq(\omega^2+\omega) + q^2)$$xyz = (p+q)(p^2 + pq(-1) + q^2)$$xyz = (p+q)(p^2 - pq + q^2)$$xyz = p^3+q^3$.

If $x=p+q$,$y=p \omega+q \omega^2$ and $z=p \omega^2+q \omega$,where $\omega$ is a complex cube root of unity,then $xyz$ is equal to

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