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

The binomial expansion is given by:

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

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

\(1 + \left(-\frac{1}{3}\right)(3x) + \frac{-\frac{1}{3}(-\frac{4}{3})}{2}(3x)^2 + \frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6}(3x)^3\)

Calculating each term:

First term: \(1\)

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

Third term: \(\frac{-\frac{1}{3} \times -\frac{4}{3}}{2} \times 9x^2 = 2x^2\)

Fourth term: \(\frac{-\frac{1}{3} \times -\frac{4}{3} \times -\frac{7}{3}}{6} \times 27x^3 = -\frac{14}{3}x^3\)

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

\(1 - x + 2x^2 - \frac{14}{3}x^3\)