Question

If $\omega$ is an imaginary cube root of unity and $\alpha$, $\beta$, $\gamma$ are the roots of real p, then for any x, y and z, the

values of $\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }$ are:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Since $\alpha$, $\beta$, $\gamma$ are cube roots of p, which are $p^{\frac{1}{3}},p^{\frac{1}{3}}\omega ,p^{\frac{1}{3}}{\omega }^2$.

Case I. Let $\alpha$ = $p^{\frac{1}{3}}$, $\beta$ = $p^{\frac{1}{3}}$$\omega$ and $\lambda$ = $p^{\frac{1}{3}}$${\omega }^2$.

Then z = $\frac{ab}{ad}$ = $\frac{ab}{ad}$

z = $\frac{ab}{ad}$ = ${\omega }^2$.

Case II. Let $\alpha$ = $p^{\frac{1}{3}}$, $\beta$ = $p^{\frac{1}{3}}$${\omega }^2$ and z = $p^{\frac{1}{3}}$$\omega$ then

z = $\frac{x+y{\omega }^2+z\omega }{x{\omega }^2+y\omega +z}$ = $\frac{\omega (x+y{\omega }^2+z\omega )}{(x+y{\omega }^2+z\omega )}$ = $\omega$

Hence Z may be $\omega$ or ${\omega }^2$. All other combination for $\alpha$, $\beta$, $\gamma$ will give same values $\omega$ or ${\omega }^2$.