The sum of cubes of two natural numbers is 280. What are the numbers if the square of its ratio is 9/4?

Let a and b be the natural
numbers:

The sum of cubes is
280:

a^3 + b^3 = 280

But the
sum of cubes returns the product:

a^3 + b^3 = (a+b)(a^2 -
ab + b^2)

The square of the ratio of a and b is
9/4.

(a/b)^2 = 9/4 => a/b =
3/2

We'll keep only the positive values, since a and b are
natural numbers.

2a = 3b => a =
3b/2

(a+b)(a^2 - ab + b^2) = (3b/2 + b)(9b^2/4 - 3b^2/2 +
b^2)

(a+b)(a^2 - ab + b^2) = (5b/2)(7b^2/4) =
35b^3/8

But (a+b)(a^2 - ab + b^2) = 280 => 35b^3/8 =
280

b^3 = 64

b = cube root
(64)

b = 4

a = 3*4/2 =
6

The requested natural numbers that respect
the imposed constraints are: a = 6 and b = 4.