Given \(\alpha + \beta + \gamma = 3\), we have \(b = - (\alpha + \beta + \gamma) = -3\).

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

\(5 = 3^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

Solving gives \(\alpha\beta + \beta\gamma + \gamma\alpha = 1\), so \(c = 2\).

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

\(6 = 3(\alpha\beta\gamma) + 3(5 - 1)\).

Solving gives \(\alpha\beta\gamma = 1\), so \(d = 1\).

The equation is \(x^3 - 3x^2 + 2x + 1 = 0\).