Question

If $\alpha$ and $\beta$ be the roots of $x^{2} + px + q = 0$, then $\frac{\left(\omega\alpha+\omega ^{2}\beta\right)\left(\omega^{2} \alpha +\omega\beta \right)}{\frac{\alpha^{2}}{\beta} + \frac{\beta^{2}}{\alpha}}$ is equal to ($\omega,\, \omega^{2}$ are complex cube roots of unity)

• A. $-\frac{q}{p}$
• B. $\alpha\,\beta$
• C. $-\frac{p}{q}$
• D. $\omega$

Answer 1

Answer: (A)

Since, $\alpha$ and $\beta$ are the roots of the equation $x^2 + px + q = 0$, therefore
$\alpha+\beta = -p$ and $\alpha\beta = q$
Now, $\left(\omega\alpha + \omega ^{2}\beta \right)\left(\omega^{ 2}\alpha + \omega \beta \right)$
$= \alpha^{2}+\beta^{2} + \left(\omega^{4}+\omega^{2}\right)\alpha\beta \quad\left(\because \omega ^{3} = 1\right)$
$= \alpha ^{2}+\beta ^{2} - \alpha \beta . \quad \left(\because \omega + \omega ^{2} = 1\right)$
$= \left(\alpha +\beta\right)^{2} -3\alpha \beta $
$= p^{2} - 3q$
Also, $\frac{\alpha^{2}}{\beta} +\frac{\beta ^{2}}{\alpha} = \frac{\alpha ^{3}+\beta ^{3}}{\alpha\beta}$
$= \frac{\left(\alpha +\beta \right)^{3} -3\alpha \beta \left(\alpha +\beta \right) }{\alpha \beta }$
$= \frac{p\left(3q-p^{2}\right)}{q}$
$\therefore \quad$ The given expression $= \frac{\left(p^{2}-3q\right)}{\frac{p\left(3q-p^{2}\right)}{q}}= -\frac{q}{p}$