Given two cubes, the volume of the first is 64 times the volume of the second. How many times is the perimeter

Given two cubes, the volume of the first is 64 times the volume of the second. How many times is the perimeter of the base of the first cube greater than the perimeter of the base of the second?

1. Let’s denote the side of the first cube as a, the side of the second cube as b.

The volume of the first cube is denoted by V1, the volume of the second cube by V2.

The perimeter of the base of the first cube is denoted by P1.

Let’s designate the perimeter of the base of the second cube behind P2.

2. Let’s calculate the volume of the first cube.

V1 = a ^ 3.

3. Let us write the equation if it is known that the volume of the second cube is 64 times the volume of the first cube.

V2 = b ^ 3 = 64 V1 = 64 a ^ 3.

From the equation, we express b through a.

b = (64 a ^ 3) = 4a.

4. Let’s write the equation for the perimeter of the base of the first cube.

P1 = a + a + a + a = 4a.

5. Let’s define the equation for the perimeter of the base of the second cube.

P2 = 4b = 4 * 4a = 16a.

6. Let’s count how many times P2 is more than P1.

P2: P1 = 16a: 4a = 4.

Answer: The perimeter of the base of the second cube is 4 times the perimeter of the base of the first cube.