Question

Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:

1. Sum of these numbers
2. Difference of their cubes
3. Sum of their cubes.

Answer 1

Given x - y = 5 and xy = 24 (x>y)

(x + y)² = (x - y)² + 4xy

(x + y)² = 25 + 96

(x + y)² = 121

(x + y) = `sqrt121`

(x + y) = ±11

So, x + y = 11; sum of these numbers is 11.

Cubing on both sides gives

(x - y)³ = 5³

x³ - y³ - 3xy(x - y) = 125

x³ - y³ - 72(5) = 125

x³ - y³ = 125 + 360

= 485

So, difference of their cubes is 485.

Cubing both sides, we get

(x + y)³ = 11³

x³ + y³ + 3xy(x + y) = 1331

x³ + y³ = 1331 - 72(11)

= 1331 - 792

= 539

So, sum of their cubes is 539.