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 need to calculate $xyz = (p+q)(p \omega+q \omega^2)(p \omega^2+q \omega)$.Fi

Vedclass · Accepted Answer

$p^3+q^3$

Vedclass · Answer

(A) Given: $x=p+q$,$y=p \omega+q \omega^2$,and $z=p \omega^2+q \omega$.We need to calculate $xyz = (p+q)(p \omega+q \omega^2)(p \omega^2+q \omega)$.First,multiply $y$ and $z$:$yz = (p \omega+q \omega^2)(p \omega^2+q \omega) = p^2 \omega^3 + pq \omega^2 + pq \omega^4 + q^2 \omega^3$.Since $\omega^3=1$ and $\omega^4=\omega$,we have:$yz = p^2(1) + pq(\omega^2+\omega) + q^2(1) = p^2 + pq(\omega^2+\omega) + q^2$.Using the property $1+\omega+\omega^2=0$,we know $\omega+\omega^2=-1$.So,$yz = p^2 - pq + q^2$.Now,$xyz = (p+q)(p^2 - pq + q^2)$.Using the algebraic identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$,we get:$xyz = p^3+q^3$.

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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