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:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.

Answer 1

Given x - y = 5 and xy = 24 (x>y)
(x + y)² = (x - y)² + 4xy = 25 + 96 = 121
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.