(a) To expand \(\sqrt[3]{1 + 6x}\), we use the binomial series expansion for \((1 + u)^n\) where \(n = \frac{1}{3}\) and \(u = 6x\).

The expansion is given by:

\(1 + \frac{n}{1!}u + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\)

Substituting \(n = \frac{1}{3}\) and \(u = 6x\), we get:

\(1 + \frac{1}{3}(6x) + \frac{1}{3} \cdot \frac{-2}{3} \cdot \frac{(6x)^2}{2} + \frac{1}{3} \cdot \frac{-2}{3} \cdot \frac{-5}{3} \cdot \frac{(6x)^3}{6}\)

Simplifying each term:

First term: \(1\)

Second term: \(2x\)

Third term: \(-4x^2\)

Fourth term: \(\frac{40}{3}x^3\)

Thus, the expansion up to \(x^3\) is \(1 + 2x - 4x^2 + \frac{40}{3}x^3\).

(b) The expansion is valid for \(|6x| < 1\), which simplifies to \(|x| < \frac{1}{6}\).