To expand \(\frac{1}{(2+x)^3}\), we can rewrite it as \((2+x)^{-3}\) and use the binomial series expansion for \((1+u)^n\), where \(u = \frac{x}{2}\) and \(n = -3\).

The binomial expansion is given by:

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

Substituting \(u = \frac{x}{2}\) and \(n = -3\), we have:

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

Simplifying each term:

First term: \(1\)

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

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

Thus, the expansion of \((1+\frac{x}{2})^{-3}\) up to \(x^2\) is:

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

Now, multiply by \(\frac{1}{8}\) to account for the factor of \(\frac{1}{8}\) in the original expression:

\(\frac{1}{8}(1 - \frac{3}{2}x + \frac{3}{4}x^2) = \frac{1}{8} - \frac{3}{16}x + \frac{3}{16}x^2\)