We are given that $$p$$ and $$q$$ are non-zero real numbers, and we have the equations $$\alpha^3 + \beta^3 = -p$$ and $$\alpha\beta = q$$. We need to find a quadratic equation with roots $$\frac{\alpha^2}{\beta}$$ and $$\frac{\beta^2}{\alpha}$$.
For a quadratic equation with roots $$r$$ and $$s$$, it can be written as $$x^2 - (r+s)x + (rs) = 0$$. So, let $$r = \frac{\alpha^2}{\beta}$$ and $$s = \frac{\beta^2}{\alpha}$$. We need to find the sum $$r+s$$ and the product $$rs$$.
First, compute the product $$rs$$:
$$$
rs = \left( \frac{\alpha^2}{\beta} \right) \times \left( \frac{\beta^2}{\alpha} \right) = \frac{\alpha^2 \beta^2}{\alpha \beta} = \alpha \beta.
$$$
Since $$\alpha\beta = q$$, we have $$rs = q$$.
Next, compute the sum $$r+s$$:
$$$
r + s = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}.
$$$
To add these fractions, find a common denominator, which is $$\alpha\beta$$:
$$$
\frac{\alpha^2}{\beta} = \frac{\alpha^2 \cdot \alpha}{\beta \cdot \alpha} = \frac{\alpha^3}{\alpha\beta}, \quad \frac{\beta^2}{\alpha} = \frac{\beta^2 \cdot \beta}{\alpha \cdot \beta} = \frac{\beta^3}{\alpha\beta}.
$$$
So,
$$$
r + s = \frac{\alpha^3}{\alpha\beta} + \frac{\beta^3}{\alpha\beta} = \frac{\alpha^3 + \beta^3}{\alpha\beta}.
$$$
Given $$\alpha^3 + \beta^3 = -p$$ and $$\alpha\beta = q$$, substitute:
$$$
r + s = \frac{-p}{q} = -\frac{p}{q}.
$$$
Now, the quadratic equation with sum of roots $$-\frac{p}{q}$$ and product of roots $$q$$ is:
$$$
x^2 - \left( -\frac{p}{q} \right)x + q = 0 \quad \rightarrow \quad x^2 + \frac{p}{q}x + q = 0.
$$$
To eliminate the fraction, multiply every term by $$q$$ (since $$q \neq 0$$):
$$$
q \cdot x^2 + q \cdot \frac{p}{q}x + q \cdot q = 0 \quad \rightarrow \quad qx^2 + px + q^2 = 0.
$$$
Comparing with the options:
A. $$px^2 - qx + p^2 = 0$$
B. $$qx^2 + px + q^2 = 0$$
C. $$px^2 + qx + p^2 = 0$$
D. $$qx^2 - px + q^2 = 0$$
The equation $$qx^2 + px + q^2 = 0$$ matches option B.
Hence, the correct answer is Option B.
91- 6303239042
