We start by expanding \(\sqrt[3]{1 - 6x}\) using the binomial series expansion for \((1 + u)^n\), where \(n = \frac{1}{3}\) and \(u = -6x\).

The binomial expansion is given by:

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

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

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

Calculating each term:

First term: \(1\)

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

Third term: \(\frac{\frac{1}{3}(-\frac{2}{3})}{2}(-6x)^2 = \frac{1}{3} \times -\frac{2}{3} \times 18x^2 = -4x^2\)

Fourth term: \(\frac{\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})}{6}(-6x)^3 = \frac{1}{3} \times -\frac{2}{3} \times -\frac{5}{3} \times -216x^3 = -\frac{40}{3}x^3\)

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

\(1 - 2x - 4x^2 - \frac{40}{3}x^3\)