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 :

(i) sum of these numbers.

(ii) difference of their cubes.

(iii) sum of their cubes.

Given,

Difference of numbers = 5 and product = 24.

Let the two numbers be x and y.

∴ x - y = 5 and xy = 24.

(i) By formula,

(x + y)² = (x - y)² + 4xy

Substituting values we get :

⇒ (x + y)² = 5² + 4 × 24

⇒ (x + y)² = 25 + 96

⇒ (x + y)² = 121

⇒ (x + y) = $\sqrt{121} = \pm 11$.

Since, numbers are positive so sum cannot be negative.

Hence, sum of these numbers = 11.

(ii) By formula,

⇒ x³ - y³ = (x - y)³ + 3xy(x - y)

⇒ x³ - y³ = 5³ + 3 × 24 × 5

⇒ x³ - y³ = 125 + 360

⇒ x³ - y³ = 485.

Hence, difference of cubes of numbers = 485.

(iii) By formula,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

Substituting x + y = 11, we get :

⇒ x³ + y³ = 11³ - 3 × 24 × 11

⇒ x³ + y³ = 1331 - 792

⇒ x³ + y³ = 539.

Hence, sum of cubes of numbers = 539.