Let p and q be real numbers such that $p \neq 0, p^{3} \neq q$ and $p^{3} \neq - q .$ If $α a n d β$ are non-zero complex numbers satisfying $α + β = - p$ and $α^{3} + β^{3} = q$ then a quadratic equation having $\frac{α}{β}$ and $\frac{β}{α}$ as its roots is

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$(p^{3} + q) x^{2} - (p^{3} + 2 q) x + (p^{3} + q) = 0$

$(p^{3} + q) x^{2} - (p^{3} - 2 q) x + (p^{3} + q) = 0$

$(p^{3} - q) x^{2} - (5 p^{3} - 2 q) x + (p^{3} - q) = 0$

$(p^{3} - q) x^{2} - (5 p^{3} + 2 q) x + (p^{3} - q) = 0$

answer is B.

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Detailed Solution

$α^{3} + β^{3} = q \Rightarrow {(α + β)}^{3} - 3 α β (α + β) = 9$
$\Rightarrow α β = \frac{q + p^{3}}{3 p}$
$x^{2} - (\frac{α}{β} + \frac{β}{α}) x + \frac{α}{β} \frac{β}{α} = 0$
$x^{2} - \frac{(α^{2} + β^{2})}{α β} x + 1 = 0$
$\Rightarrow x^{2} - \frac{({(α + β)}^{2} - 2 α β)}{α β} x + 1 = 0$
$\Rightarrow x^{2} - \frac{p^{2} - 2 (\frac{p^{3} + q}{3 p})}{\frac{p^{3} + q}{3 p}} x + 1 = 0$
$\Rightarrow (p^{3} + q) x^{2} - (p^{3} - 2 q) x + (p^{3} + q) = 0$