When calculating the volume, raise the power of the radius to 3. This means that relative to to the radius, its volume will also raise to the power of 3
If we double the radius of the sphere, then
[tex]\begin{gathered} \text{If} \\ r_1=3 \\ r_2=3\cdot2=6 \\ \\ \text{ Volume of Sphere 1} \\ r=3 \\ V=\frac{4}{3}\pi r^3 \\ V=\frac{4}{3}\pi(3)^3 \\ V=\frac{4}{3}\pi(27) \\ V=36\pi \\ \\ \text{Volume of Sphere 2} \\ V=\frac{4}{3}\pi r^3 \\ V=\frac{4}{3}\pi(6)^3 \\ V=\frac{4}{3}\pi(216) \\ V=288\pi \end{gathered}[/tex]Comparing the two volumes
[tex]\frac{288\pi}{36\pi}=8[/tex]Therefore, the volume of the new sphere is 8 times as much.