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

x - y = 5 ----> eq 1

xy = 24

y = 24/x ----> substitute in eq 1

x - 24/x = 5

(x² - 24)/x = 5

x² - 24 = 5x

x² - 5x - 24 = 0

x² - 8x + 3x - 24 = 0

x(x - 8) + 3(x - 8) = 0

(x + 3)(x - 8) = 0

(x + 3) = 0 or (x-8) = 0

x = -3 or x = 8

y = -8 or y = 3

(i) x + y = -11 or +11

(ii) x³ - y³ = -27 - (-512) or 512 - 27 = 485

(iii) x³ + y³ = - 27 - 512 or 512 + 27 = - 539 or 539

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Answer 2

given x-y=5

xy=24

(x+y) ^2 = (x-y)^2 +4xy

(x+y)^2 = (5)^2 + 4(24)

(x+y)^2 = 25 + 96

(x+y)^2 = 121

x+y = 11

sum of these no. is 11

x-y = 5

cube both side

(x-y)^3 = (5)^3

x^3 -y^3 -3xy(x-y) = 125

X3 -y3 -3(24)(5) = 125

x3-y3 = 125+360

x3-y3 = 485

difference of their cubes is 485

x+y =11

cube both side

(x+y)^3 = (11)^3

x^3 +y^3 +3xy(x+y) = 1331

X3 +y3 +3(24)(11) = 1331

x3+y3 = 1331-729

x3+y3 = 539

sum of their cubes is 539

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