The sum of the cubes of two given natural numbers is 9728, while the sum of the two given numbers is 32. What is the positive difference between the cubes of the two given numbers?

To find the positive difference between the cubes of the two numbers, we can first determine the numbers themselves by using algebraic identities.

Let the two natural numbers be $x$ and $y$.

From the problem, we are given two conditions:
1. The sum of the two numbers is 32:

$x+y=32$
2. The sum of the cubes of the two numbers is 9728:

$x^3+y^3=9728$

We can use the algebraic identity for the sum of cubes:

$x^3+y^3=(x+y)({(x+y)}^2-3xy)$

Substitute the given values into this identity:

$9728=32·({32}^2-3xy)$

Divide both sides of the equation by 32:

$304=1024-3xy$

Rearrange the terms to solve for $xy$:

$3xy=1024-304$

$3xy=720$

$xy=240$

Now we know that the sum of the two numbers is 32 and their product is 240. We can set up a quadratic equation to find the two numbers:

$t^2-32t+240=0$

Factoring the quadratic equation:

$(t-20)(t-12)=0$

This gives the roots as:

$t=20$ or $t=12$
So, the two natural numbers are $x=20$ and $y=12$.

Now, we calculate the cubes of these two numbers:

${20}^3=8000$

${12}^3=1728$

Finally, we find the positive difference between the cubes:

$8000-1728=6272$

The correct answer is 6272.