(a) Using the formula for the sum of squares of roots, \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

From the cubic equation, \(\alpha + \beta + \gamma = -2\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 3\).

Thus, \((-2)^2 - 2(3) = 4 - 6 = -2\).

(b) Using the formula for the sum of cubes of roots, \(\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha\beta\gamma) + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)\).

\(\alpha\beta\gamma = -3\), so \(\alpha^3 + \beta^3 + \gamma^3 = 3(-3) + (-2)(-2 - 3) = -9 + 10 = 1\).

(c) Expanding \((\alpha + r)^3 = \alpha^3 + 3\alpha^2 r + 3\alpha r^2 + r^3\).

Thus, \(\sum_{r=1}^{n} ((\alpha + r)^3 + (\beta + r)^3 + (\gamma + r)^3) = \sum_{r=1}^{n} (1 + 3(-2)r + 3(-2)r^2 + 3r^3)\).

Using standard results, \(n - 6(\frac{1}{2}n(n+1)) - 6(\frac{1}{6}n(n+1)(2n+1)) + \frac{3}{4}n^2(n+1)^2\).

Simplifying gives \(n + \frac{1}{4}n(n+1)(-12 - 4(2n+1) + 3n(n+1))\).

Finally, \(n + \frac{1}{4}n(n+1)(3n^2 - 5n - 16)\).