(a) The determinant of the matrix \(\begin{pmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{pmatrix}\) is calculated as:

\(\left| \begin{array}{ccc} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} 1 & \gamma \\ \gamma & 1 \end{array} \right| - \alpha \cdot \left| \begin{array}{cc} \alpha & \gamma \\ \beta & 1 \end{array} \right| + \beta \cdot \left| \begin{array}{cc} \alpha & 1 \\ \beta & \gamma \end{array} \right|\)

\(= 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\)

Since the matrix is singular, the determinant is zero:

\(1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma = 0\)

Given \(\alpha\beta\gamma = 1\), we have:

\(\alpha^2 + \beta^2 + \gamma^2 = 1 + 2(1) = 3\)

(b) We use the identity:

\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)

Given \(\alpha^2 + \beta^2 + \gamma^2 = 3\), we have:

\(3 = b^2 - 2c\)

Also, given \(\alpha^3 + \beta^3 + \gamma^3 = 3\), we use:

\(\alpha^3 + \beta^3 + \gamma^3 = -b(3) - c(-b) + 3\)

\(3 = -3b + cb + 3\)

Solving these simultaneous equations, we find:

\(c = 3, \ b = 3\)