Let $p$ and $q$ be real numbers such that $p \neq 0, p^3 \neq q$ and $p^3 \neq-q$. If $\alpha$ and $\beta$…
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 non-zero 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
$\left(p^3+q\right) x^2-\left(p^3+2 q\right) x +\left(p^3+q\right)=0$
$\left(p^3+q\right) x^2-\left(p^3-2 q\right) x +\left(p^3+q\right)=0$
$\left(p^3-q\right) x^2-\left(5 p^3-2 q\right) x +\left(p^3-q\right)=0$
$\left(p^3-q\right) x^2-\left(5 p^3+2 q\right) x +\left(p^3-q\right)=0$
Solution
Sum of roots $=\frac{\alpha^2+\beta^2}{\alpha \beta}$ and product $=1$
Given, $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$
$
\begin{aligned}
& \Rightarrow(\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)=q \\
& \therefore \quad \alpha^2+\beta^2-\alpha \beta=\frac{-q}{p}
\end{aligned}
$
and $\quad(\alpha+\beta)^2=p^2$
$
\Rightarrow \quad \alpha^2+\beta^2+2 \alpha \beta=p^2
$
From Eqs. (i) and (ii), we get
$
\alpha^2+\beta^2=\frac{p^3-2 q}{3 p}
$
and $\alpha \beta=\frac{p^3+q}{3 p}$
$\therefore$ Required equation is
$
\begin{gathered}
x^2-\frac{\left(p^3-2 q\right) x}{\left(p^3+q\right)}+1=0 \\
\Rightarrow\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0
\end{gathered}
$