We need to expand \((1 + 6x)^{-\frac{1}{3}}\) using the binomial series expansion for negative exponents.

The binomial series for \((1 + u)^n\) is given by:

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

Here, \(n = -\frac{1}{3}\) and \(u = 6x\).

First term: \(1\)

Second term: \(-\frac{1}{3} \times 6x = -2x\)

Third term: \(\frac{-\frac{1}{3}(-\frac{4}{3})}{2} \times (6x)^2 = 8x^2\)

Fourth term: \(\frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6} \times (6x)^3 = -\frac{112}{3}x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 - 2x + 8x^2 - \frac{112}{3}x^3\)