Let the first term be denoted as \(a = 216\) and the common ratio as \(r\).

The fourth term is given by \(ar^3 = 64\).

Substituting the known value of \(a\), we have:

\(216r^3 = 64\)

Solving for \(r\), we get:

\(r^3 = \frac{64}{216} = \frac{8}{27}\)

\(r = \left(\frac{8}{27}\right)^{1/3} = \frac{2}{3}\)

The sum to infinity of a geometric progression is given by:

\(S_\infty = \frac{a}{1-r}\)

Substituting the values of \(a\) and \(r\), we have:

\(S_\infty = \frac{216}{1 - \frac{2}{3}} = \frac{216}{\frac{1}{3}} = 648\)