The cubes strictly between the limits are \(5^3=125,6^3=216,7^3=343,8^3=512,9^3=729\).

The first four lie strictly between consecutive squares: \(11^2\lt 125\lt 12^2\), \(14^2\lt 216\lt 15^2\), \(18^2\lt 343\lt 19^2\), and \(22^2\lt 512\lt 23^2\).

The remaining cube is \(729=27^2\), so it works. Cubes outside the list are outside the interval, which proves uniqueness.

Answer

The number is \(729\), with \(\sqrt{729}=27\) and \(\sqrt[3]{729}=9\).