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# help

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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 $$a+b$$.

Mar 28, 2019

$$\dfrac 4 3 \pi \left(\dfrac D 2\right)^3 = 2V=\dfrac 8 3 \pi \left(\dfrac{9}{2}\right)^3\\ \left(\dfrac D 2\right)^3 = 2\left(\dfrac 9 2 \right)^3 = \left(\sqrt[3]{2}~\dfrac 9 2 \right)^3\\ D = 9\sqrt[3]{2}\\ 9+2=11$$