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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Current $I$ flows through a long conducting wire bent at right angles as shown in the figure. The magnetic field at a point $P$ on the right bisector of the angle $XOY$ at a distance $r$ from $O$ is

The area (in sq units) of the quadrilateral formed by two pairs of lines $\lambda^2 x^2 - m^2 y^2 - n(\lambda x + m y) = 0$ and $\lambda^2 x^2 - m^2 y^2 + n(\lambda x + m y) = 0$ is:

If $\int(1-\cos x) \operatorname{cosec}^2 x \, dx = f(x) + c$, then $f(x)$ is equal to

With a standard rectangular bar magnet,the time period of a vibration magnetometer is $4 ~s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module