Question

Let x and y be positive real numbers such that x3-y3 +(x - y)^3 - 36xy = 3456. Find the value of x-y.​

Answer 1

Given : x³ - y³   + (x - y)³  - 36xy  = 3456

To find :    the value of x-y.​

Solution:

x³ - y³   + (x - y)³  - 36xy  = 3456  

Let say x - y = a     * xy =  b

x³ - y³ = (x - y)³  + 3xy(x - y)  = a³ + 3ab

a³ + 3ab   + a³  -36b = 3456

=> 2a³  - 3456  + 3ab - 36b = 0

=> 2(a³ - 1728)  + 3b(a  - 12)  = 0

=>  2(a³ - 12³)  + 3b(a  - 12)  = 0

=> 2((a - 12) (a² + 12² + 24a)   + 3b(a  - 12)  = 0

=> (a - 12) ( 2a² + 288 + 48a - 3b) = 0

=> a - 12 = 0

=> a = 12

=> x - y = 12

x - y = 12  

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