If alpha, beta, gamma are the cube roots of p | vedclass

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

Question

(D) Let $p = -q$ where $q > 0$. The cube roots of $p$ are $\alpha = -q^{1/3}$,$\beta = -q^{1/3}\omega$,and $\gamma = -q^{1/3}\omega^2$,where $\omega =

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $p = -q$ where $q > 0$. The cube roots of $p$ are $\alpha = -q^{1/3}$,$\beta = -q^{1/3}\omega$,and $\gamma = -q^{1/3}\omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ is the complex cube root of unity.Substituting these into the expression:$\frac{x(-q^{1/3}) + y(-q^{1/3}\omega) + z(-q^{1/3}\omega^2)}{x(-q^{1/3}\omega) + y(-q^{1/3}\omega^2) + z(-q^{1/3}\omega^3)} = \frac{-(x + y\omega + z\omega^2)}{-\omega(x + y\omega + z\omega^2)} = \frac{1}{\omega} = \omega^2$.Since $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$,none of the given options match.

If $\alpha, \beta, \gamma$ are the cube roots of $p$ $(p < 0)$,then for any $x, y$ and $z$,$\frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha} = $

Similar Questions

One of the roots of the equation $(x+1)^4 + 81 = 0$ is

$A$ complex number $z$ among the following which satisfies $z^3+27 i=0$ is

If $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i}$,then $(\bar{z})^{100}$ lies in

Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, where $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is . . . . . . .

If the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 1)^3 + 8 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module