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 $a+b$.
To solve this problem, I need to find a relationship between the volumes and the radii of the two spheres. I recall that the volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$, where $r$ is the radius. If I let $V_1$ and $V_2$ be the volumes of the smaller and larger spheres, respectively, and $r_1$ and $r_2$ be their radii, then I have $V_1 = \frac{4}{3} \pi r_1^3$ and $V_2 = \frac{4}{3} \pi r_2^3$. I also know that $V_2 = 2V_1$, so I can substitute this into the equation for $V_2$ and get $\frac{4}{3} \pi r_2^3 = 2 \frac{4}{3} \pi r_1^3$. Simplifying this equation, I get $r_2^3 = 2r_1^3$. I can plug in $r_1 = 9$ and get $r_2^3 = 2 \cdot 9^3 = 1458$. To find $r_2$, I need to take the cube root of both sides of the equation. I notice that $1458$ has a factor of $2$ and a factor of $729$, which is $9^3$. So I can rewrite $r_2^3 = 2 \cdot 9^3 = 2 \cdot 729 = 2 \cdot (9 \cdot 9 \cdot 9)$. Taking the cube root of this, I get $r_2 = \sqrt[3]{2} \cdot 9$. This means that the diameter of the larger sphere is $2r_2 = 2 \sqrt[3]{2} \cdot 9$. This is of the form $a\sqrt[3]{b}$, where $a = 18$ and $b = 2$. Therefore, the answer is $a+b = 18+2 = 20$. 