(a) Let \(y = x^3\). Then \(y^{4/3} - 2y^{1/3} - 1 = 0\). Substitute \(y = x^3\) into the equation to get \(y^4 - 8y^3 - 12y^2 - 6y - 1 = 0\). The value of \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3\) is 8.

(b) The expression \(\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} + \frac{1}{\delta^3}\) can be rewritten using the identity \(\frac{\alpha^3 \beta^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \gamma^3 + \beta^3 \gamma^3 \delta^3 + \alpha^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \delta^3}{\alpha^3 \beta^3 \gamma^3 \delta^3} = \frac{6}{-1} = -6\).

(c) Using the original equation, \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 2(\alpha^3 + \beta^3 + \gamma^3 + \delta^3) + 4 = 2(8) + 4 = 20\).