Let V1 be the volume of the large sphere:
[tex]V_1=\frac{4}{3}\pi\cdot r^3_l_{}[/tex]and let V2 be the volume of the smaller sphere:
[tex]V_2=\frac{4}{3}\pi\cdot r^3_s[/tex]since the radius of the large sphere is twice the radius of the small sphere, we have the following equation:
[tex]r_l=2r_s[/tex]if we do this substitution on the first equation, we get the following:
[tex]\begin{gathered} V_1=\frac{4}{3}\pi(2r_s)^3=\frac{4}{3}\pi\cdot8r^3_s=8\cdot(\frac{4}{3}\pi\cdot r^3_s)=8\cdot V_2 \\ \Rightarrow V_1=8\cdot V_2_{} \end{gathered}[/tex]therefore, the volume of the larger sphere is 8 times the volume of the smaller sphere
