To expand \((1 - 4x)^{\frac{1}{4}}\), we use the binomial series expansion for \((1 + u)^n\), which is:

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

Here, \(n = \frac{1}{4}\) and \(u = -4x\).

The first term is \(1\).

The second term is \(\frac{1}{4}(-4x) = -x\).

The third term is \(\frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{(-4x)^2}{2} = -\frac{3}{2}x^2\).

The fourth term is \(\frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{-7}{4} \cdot \frac{(-4x)^3}{6} = -\frac{7}{2}x^3\).

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

\(1 - x - \frac{3}{2}x^2 - \frac{7}{2}x^3\).