The dimensions of the cube were enlarged, respectively, 2 times, 4 times and 8 times and received a rectangular parallelepiped.

The dimensions of the cube were enlarged, respectively, 2 times, 4 times and 8 times and received a rectangular parallelepiped. How many times is the volume of a rectangular parallelepiped larger than the volume of a cube?

Let’s denote x – the initial volume of the cube; V is the volume of the parallelepiped.
Let x = a * a * a, where a is the length, width, or height of the cube. Then, by the condition of the problem, V = (a * 2) * (a * 4) * (a * 8) = x * 64. That is, the volume of the new parallelepiped is 64 times the original volume of the cube.