(a) Let \(y = 3x + 1\), then \(x = \frac{1}{3}(y - 1)\).

Substitute into the original equation:

\(27\left(\frac{y-1}{3}\right)^3 + 18\left(\frac{y-1}{3}\right)^2 + 6\left(\frac{y-1}{3}\right) - 1 = 0\)

\(\Rightarrow (y-1)^3 + 2(y-1)^2 + 2(y-1) - 1 = 0\)

\(\Rightarrow y^3 - 3y^2 + 3y - 1 + 2y^2 - 4y + 2 + 2y - 2 - 1 = 0\)

\(\Rightarrow y^3 - y^2 + y - 2 = 0\)

(b) \(S_2 = 1^2 - 2(1) = -1\)

\(S_3 = (3\alpha + 1)^3 + (3\beta + 1)^3 + (3\gamma + 1)^3 = -1 - (1) + 6 = 4\)

(c) \(S_{-1} = \frac{(3\alpha + 1)(3\beta + 1) + (3\beta + 1)(3\gamma + 1) + (3\gamma + 1)(3\alpha + 1)}{(3\alpha + 1)(3\beta + 1)(3\gamma + 1)} = \frac{1}{2}\)

\(2S_{-2} = S_1 + S_{-1} = 1 - 3 + \frac{1}{2}\)

\(S_{-2} = -\frac{3}{4}\)