Respuesta :
just a quick addition to the great reply by yoyotam526 above.
[tex]~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\\begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{model}{1}~:~\stackrel{actual}{9}~~\implies \cfrac{\stackrel{model}{1}}{\underset{actual}{9}}=\cfrac{\sqrt[3]{\stackrel{model}{V}}}{\sqrt[3]{\stackrel{actual}{V}}}\implies \cfrac{1}{9}=\cfrac{\sqrt[3]{V}}{\sqrt[3]{5770}}\implies \cfrac{1}{9}=\sqrt[3]{\cfrac{V}{5770}} \\\\\\ \left( \cfrac{1}{9} \right)^3=\cfrac{V}{5770}\implies \cfrac{1}{9^3}=\cfrac{V}{5770}\implies \cfrac{1}{729}=\cfrac{V}{5770} \\\\\\ \cfrac{5770}{729}=V\implies 7.915~\approx~V\implies \stackrel{\textit{rounded up}}{8 \approx V}[/tex]