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

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Let the two numbers be A and B.

The sum of cubes of the numbers is 280.

=> A^3 + B^3 = 280 ...(1)

The square of their ratio is 9/4

=> (A/B)^2 = 9/4

=> A/B = 3/2

=> A = 3*B/2

Substitute in (1)

=> (3*B/2)^3 + B^3 = 280

=> B^3*[27/8 + 1] = 280

=> B^3*[35/8] = 280

=> B^3 = 280*8/35

=> B^3 = 64

=> B = 4

A = 3*B/2 = 3*4/2 = 6

**The two numbers are 4 and 6**

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.**