Let the three numbers be $x$, $2x$, and $3x$, where $x$ is a common factor.

The sum of their cubes is given as 62208:

$x^3 + (2x)^3 + (3x)^3 = 62208$

Expanding the cubes:

$x^3 + 8x^3 + 27x^3 = 62208$

Simplifying:

$36x^3 = 62208$

Solving for $x^3$:

$x^3 = \frac{62208}{36} = 1728$

Taking the cube root of both sides:

$x = \sqrt[3]{1728} = 12$

Thus, the three numbers are:

$x = 12, \quad 2x = 24, \quad 3x = 36$

The numbers are 12, 24, and 36.

Would you like more details on this solution or have any other questions?

Here are 5 related questions:

1. What is the sum of the three numbers?
2. How do you find the ratio of three numbers?
3. How would the solution change if the sum of cubes were different?
4. Can you solve for numbers if the ratio was 2:3:4 instead?
5. How would you verify the solution using another method?

Tip: Cube roots can be tricky. Use estimation or a calculator for precise results!