To expand \((1 + 2x)^{-3}\), we use the binomial series expansion for negative exponents:

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

Here, \(u = 2x\) and \(n = -3\).

First term: \(1\).

Second term: \(-3(2x) = -6x\).

Third term: \(\frac{-3(-4)}{2}(2x)^2 = \frac{12}{2}(4x^2) = 24x^2\).

Thus, the expansion up to \(x^2\) is \(1 - 6x + 24x^2\).