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

Given \(\alpha + \beta + \gamma = -4\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 6\), we have:

\((-4)^2 - 2(6) = 16 - 12 = 4\).

(b) Expand \((\alpha + r)^2 = \alpha^2 + 2\alpha r + r^2\).

Thus, \(\sum_{r=1}^{n} ((\alpha + r)^2 + (\beta + r)^2 + (\gamma + r)^2) = \sum_{r=1}^{n} (4 + 2(-4)r + 3r^2)\).

Calculate: \(4n - 4n(n+1) + \frac{1}{2}n(n+1)(2n+1)\).

Simplify: \(-4n^2 + \frac{1}{2}n(n+1)(2n+1)\).

\(= n(-4n + \frac{1}{2}(2n^2 + 3n + 1))\).

\(= n(n^2 - \frac{5}{2}n + \frac{1}{2})\).