Let p and q real number such that `p!= 0`,`p^3!=q` and `p^3!=-q`. if `alpha` and `beta` are non-zero complex number satisfying `alpha+beta=-p` and `alpha^3+beta^3=q`, then a quadratic equation having `alpha/beta` and `beta/alpha` as its roots is

Options

A. `(p^(3)+q)x^(2)-(p^(3)+2q)x+(p^(3)+q)=0`
B. `(p^(3)+q)x^(2)-(p^(3)-2q)x+(p^(3)+q)=0`
C. `(p^(3)-q)x^(2)-(5p^(3)-2q)x+(p^(3)-q)=0`
D. `(p^(3)-q)x^(2)(5p^(3)+2q)x+(p^(3)-q)=0`

Explanation

To solve the given problem step by step, we will derive the quadratic equation having \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) as its roots.

### Step 1: Use the given equations
We start with the equations provided:
1. \(\alpha + \beta = -p\)
2. \(\alpha^3 + \beta^3 = q\)

### Step 2: Express \(\alpha^3 + \beta^3\) in terms of \(\alpha + \beta\) and \(\alpha \beta\)
Using the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2) \]
We can rewrite \(\alpha^2 - \alpha \beta + \beta^2\) using \((\alpha + \beta)^2\):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Thus,
\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) \]
Substituting \(\alpha + \beta = -p\):
\[ \alpha^3 + \beta^3 = -p((-p)^2 - 3\alpha\beta) = -p(p^2 - 3\alpha\beta) \]
Setting this equal to \(q\):
\[ -p(p^2 - 3\alpha\beta) = q \]

### Step 3: Solve for \(\alpha \beta\)
Rearranging gives:
\[ p^3 - 3p\alpha\beta = -q \]
Thus,
\[ 3p\alpha\beta = p^3 + q \]
So,
\[ \alpha\beta = \frac{p^3 + q}{3p} \]

### Step 4: Find the roots \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\)
The roots of the quadratic equation can be expressed as:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} \]
Using the earlier derived expression for \(\alpha^2 + \beta^2\):
\[ \alpha^2 + \beta^2 = (-p)^2 - 2\alpha\beta = p^2 - 2\left(\frac{p^3 + q}{3p}\right) \]
Substituting \(\alpha\beta\):
\[ \alpha^2 + \beta^2 = p^2 - \frac{2(p^3 + q)}{3p} = p^2 - \frac{2p^2 + 2q}{3} = \frac{3p^2 - 2p^2 - 2q}{3} = \frac{p^2 - 2q}{3} \]
Thus,
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{p^2 - 2q}{3}}{\frac{p^3 + q}{3p}} = \frac{p(p^2 - 2q)}{p^3 + q} \]

### Step 5: Form the quadratic equation
The quadratic equation with roots \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) can be written as:
\[ x^2 - \left(\frac{p(p^2 - 2q)}{p^3 + q}\right)x + 1 = 0 \]

### Final Result
The required quadratic equation is:
\[ x^2 - \frac{p(p^2 - 2q)}{p^3 + q}x + 1 = 0 \]