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

Let p and q be real numbers such that p ≠ 0, p³ ≠ q and p³ ≠ -q. If α and β are nonzero complex numbers satisfying α + β = -p and α³ + β³ = q, then a quadratic equation having α/β and β/α as its roots is

$\text{Given: } p,q \in \mathbb{R},\quad p \neq 0,\quad p^{3} \neq q,\quad p^{3} \neq -q$

$\alpha,\beta \text{ nonzero complex},\quad \alpha+\beta=-p$

$\alpha^{3}+\beta^{3}=q$

$\text{To find: quadratic equation with roots } \frac{\alpha}{\beta} \text{ and } \frac{\beta}{\alpha}$

$\text{Plan: } \alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)$

$\text{Need } \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \text{ and } \frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha}$

$s=\alpha+\beta=-p,\qquad t=\alpha\beta$

$\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)$

$=s^{3}-3ts$

$q=s^{3}-3ts=(-p)^{3}-3t(-p)$

$q=-p^{3}+3pt$

$3pt=p^{3}+q \ \Rightarrow\ t=\frac{p^{3}+q}{3p}$

$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha\beta}$

$\frac{\alpha^{2}+\beta^{2}}{\alpha\beta}=\frac{s^{2}-2t}{t}=\frac{p^{2}}{t}-2$

$\frac{p^{2}}{t}=\frac{p^{2}\cdot 3p}{p^{3}+q}=\frac{3p^{3}}{p^{3}+q}$

$\frac{3p^{3}}{p^{3}+q}-2=\frac{3p^{3}-2p^{3}-2q}{p^{3}+q}$

$=\frac{p^{3}-2q}{p^{3}+q}$

$\frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha}=\frac{\alpha\beta}{\alpha\beta}=1$

$x^{2}-\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)x+\frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha}=0$

$x^{2}-\frac{p^{3}-2q}{p^{3}+q}\,x+1=0$

$(p^{3}+q)\left(x^{2}-\frac{p^{3}-2q}{p^{3}+q}x+1\right)=0$

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

$\text{Check: } p=1,\ q=2 \Rightarrow p^{3}-2q=-3,\ p^{3}+q=3$

$3x^{2}+3x+3=0 \Rightarrow x^{2}+x+1=0$

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

$\text{Option (2)}$