Simplify the given expression.

$\frac{[120\times120\times120-100\times100\times100]}{[120\times120+120\times100+100\times100]}$

If (a + b + c) = 14, and (a³ + b³ + c³ - 3abc) = 98, find the value of (a² + b² + c²).

The difference between the cubes of two given natural numbers is 6272 , while the positive difference between the two given numbers is 8. What is the sum of the cubes of the two given numbers?

If 2x + 3y = 16 and xy = 9, then find 8x³ + 27y³