To solve this problem, we need to understand the nature of the numbers involved. The cube roots of a number are solutions to the equation obtained by setting that number equal to \( x^3 \). Hence, we begin by recognizing that the cube roots of \(-8\) satisfy \( x^3 = -8 \).

Let's express -8 in polar form: \(-8 = 8 \text{cis} 180^\circ\) Taking the cube root, we have: \(\alpha = 2 \text{cis} (60^\circ), \quad \beta = 2 \text{cis} (180^\circ), \quad \gamma = 2 \text{cis} (300^\circ)\)

These roots form an equilateral triangle in the complex plane and satisfy: \(\alpha^3 = \beta^3 = \gamma^3 = -8 \quad \text{and} \quad \alpha \beta \gamma = -8\)

Given the expression: \(\frac{\alpha^2 p^2 + \beta^2 q^2 + \gamma^2 r^2}{\beta^2 p^2 + \gamma^2 q^2 + \alpha^2 r^2}\) It is often helpful to evaluate components like symmetry or factors in the roots to simplify.

Notice that: \(\alpha^2 = 4 \text{cis}(120^\circ), \quad \beta^2 = 4 \text{cis}(240^\circ), \quad \gamma^2 = 4 \text{cis}(0^\circ)\)

Therefore, the given expression: \(\frac{4 \text{cis}(120^\circ) p^2 + 4 \text{cis}(240^\circ) q^2 + 4 \text{cis}(0^\circ) r^2}{4 \text{cis}(240^\circ) p^2 + 4 \text{cis}(0^\circ) q^2 + 4 \text{cis}(120^\circ) r^2}\) simplifies (since \(\alpha^2, \beta^2, \gamma^2\) are symmetric) to: \(\frac{\gamma}{\beta}\)

Thus, the correct answer is: \(\frac{\gamma}{\beta}\).