Construct Quadratic Equation from Ratio of Symmetric Complex Roots

To find a quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$, we first determine the sum and the product of these roots.

Let the roots be $r_1 = \frac{\alpha}{\beta}$ and $r_2 = \frac{\beta}{\alpha}$.

The sum of the roots is:
$S = r_1 + r_2 = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$

The product of the roots is:
$P = r_1 \cdot r_2 = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$

We are given:
$\alpha + \beta = -p$
$\alpha^3 + \beta^3 = q$

Using the algebraic identity for the sum of cubes:
$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$

Substitute the given values into the identity:
$q = (-p)^3 - 3\alpha\beta(-p)$
$q = -p^3 + 3p\alpha\beta$

Since $p \neq 0$, we can solve for the product $\alpha\beta$:
$3p\alpha\beta = p^3 + q \implies \alpha\beta = \frac{p^3 + q}{3p}$

Next, we express $\alpha^2 + \beta^2$ in terms of $p$ and $q$:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$\alpha^2 + \beta^2 = (-p)^2 - 2\left(\frac{p^3 + q}{3p}\right) = p^2 - \frac{2(p^3 + q)}{3p}$
$\alpha^2 + \beta^2 = \frac{3p^3 - 2p^3 - 2q}{3p} = \frac{p^3 - 2q}{3p}$

Now, substitute $\alpha^2 + \beta^2$ and $\alpha\beta$ into the expression for $S$:
$S = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{\frac{p^3 - 2q}{3p}}{\frac{p^3 + q}{3p}} = \frac{p^3 - 2q}{p^3 + q}$

The quadratic equation with roots $r_1$ and $r_2$ is given by:
$x^2 - Sx + P = 0$
$x^2 - \left(\frac{p^3 - 2q}{p^3 + q}\right)x + 1 = 0$

Since $p^3 \neq -q$, we can multiply the entire equation by $(p^3 + q)$:
$(p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0$

Therefore, the correct option is B.