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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.
Solution
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.
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