The diameter, in inches, of a sphere with twice the volume of a sphere of radius 9 inches can be expressed in the form \(a\sqrt[3]{b}\) where \(a\) and \(b\) are positive integers and \(b\) contains no perfect cube factors. Compute \(a+b\).
\(\dfrac 4 3 \pi \left(\dfrac D 2\right)^3 = 2V=\dfrac 8 3 \pi \left(\dfrac{9}{2}\right)^3\\ \left(\dfrac D 2\right)^3 = 2\left(\dfrac 9 2 \right)^3 = \left(\sqrt[3]{2}~\dfrac 9 2 \right)^3\\ D = 9\sqrt[3]{2}\\ 9+2=11 \)
